Integrals: Chapter 7 Links
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Integrals 7.7 is a continuation of Exercise 7.4. In that exercise, we learned six important special integral formulas involving quadratic expressions. In this exercise, we will study three more standard integrals in which the square root of a quadratic expression appears in the numerator.
\displaystyle \int \sqrt{a^2-x^2}\,dx,\quad \int \sqrt{x^2\pm a^2}\,dx
These integrals are extensions of the standard forms studied earlier. Instead of integrating the reciprocal of the square root, we now integrate the square root itself. The required formulas can then be applied directly after simplifying the given expression whenever necessary.
Most questions in this exercise can be solved by first reducing the integrand to one of these three standard forms using suitable algebraic simplification or substitution. Therefore, remembering these formulas and recognizing the correct pattern is the key to solving the problems quickly.
Key Concepts
1. Three More Special Integrals
This exercise introduces three more standard integral formulas involving the square root of quadratic expressions.
These formulas are natural extensions of the last three special integrals studied in Exercise 7.4. Once the integrand matches one of these standard forms, the required formula can be applied directly.
2. Recognising Standard Forms
The most important step in this exercise is identifying whether the given integral can be reduced to one of the following standard forms.
- \sqrt{a^2-x^2}
- \sqrt{x^2-a^2}
- \sqrt{x^2+a^2}
- Expressions that can be converted into one of these forms after completing the square or using a suitable substitution.
3. Completing the Square
When a quadratic expression does not immediately match a standard formula, rewrite it by completing the square.
For example,
x^2-6x+13=(x-3)^2+4
This helps convert the integral into one of the three standard forms listed above.
4. Integration by Parts
The three special integral formulas in this exercise can also be derived using Integration by Parts, taking 1 as the second function. Although these derivations are not usually required while solving NCERT questions, they help in understanding and remembering the formulas.
5. Strategy for Solving Questions
- Identify the standard form hidden in the integrand.
- Complete the square whenever required.
- Use a suitable substitution, if necessary.
- Reduce the integral to one of the three special integral formulas.
- Apply the formula and add the constant of integration C.
3 More Special Integral Formulas
These three standard results are extensions of Formulas 4, 5 and 6 from Exercise 7.4. They can be obtained either by using the same trigonometric substitutions as before or by applying Integration by Parts, taking 1 as the second function.
| 3 More Special Integral Formulas |
|---|
| 1. \displaystyle \int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C |
| T.P. Put x=a\sin\theta (or x=a\cos\theta), same as Formula 4 of Exercise 7.4, or use Integration by Parts taking 1 as the second function. |
| 2. \displaystyle \int \sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C |
| T.P. Put x=a\sec\theta (or x=a\cosec\theta), same as Formula 5 of Exercise 7.4, or use Integration by Parts taking 1 as the second function. |
| 3. \displaystyle \int \sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C |
| T.P. Put x=a\tan\theta (or x=a\cot\theta), same as Formula 6 of Exercise 7.4, or use Integration by Parts taking 1 as the second function. |
Common Integral Forms
Most questions in this exercise involve the square root of a quadratic expression. The main objective is to reduce the given integral to one of the three standard forms listed above.
| 2 Common Integral Forms |
|---|
| 1. \displaystyle \int \sqrt{\text{Quadratic}}\,dx e.g. \displaystyle \int \sqrt{ax^2+bx+c}\,dx |
| Method: Complete the square inside the square root and reduce the integral to one of the three standard forms listed above. |
| 2. \displaystyle \int (\text{Linear})\sqrt{\text{Quadratic}}\,dx e.g. \displaystyle \int (px+q)\sqrt{ax^2+bx+c}\,dx |
| Method: Find constants A and B such that \text{Linear}=A(\text{Derivative of Quadratic})+B e.g. px+q=A(2ax+b)+B Split the integral into two parts, say I_1 and I_2. Then I_1 can be evaluated by substitution, while I_2 reduces to Type 1 above. |
Curious Thought? In Exercise 7.7, we study the integrals of \sqrt{a^2-x^2},\ \sqrt{x^2-a^2},\ \sqrt{x^2+a^2}, which are the “opposite” of the last three special integrals from Exercise 7.4. You may wonder about the reciprocals of the first three formulas of Exercise 7.4. Interestingly, those are simply standard integrals that can be evaluated using the usual integration formulas like power rule, so no separate special formulas are required!
Let us now solve all the NCERT questions step by step in this exercise 7.7 of Integrals.
Question 1: Integrals 7.7
1. Evaluate \displaystyle \int \sqrt{4-x^2}\,dx.
Important: This question can be solved in three different ways:
- Method 1: Using the standard integral formula (recommended for examinations).
- Method 2: Using the trigonometric substitution x=2\sin\theta, which was introduced in Exercise 7.2 and leads to the same result.
- Method 3: Using Integration by Parts, from which the standard formula itself can be derived.
In the remaining questions of this exercise, we shall use the standard formulas only, as they are the quickest and most efficient approach after reducing the integral to the required standard form.
Method 1: Using the Standard Formula
Solution
Let, \displaystyle I=\int \sqrt{4-x^2}\,dx
Rewrite the expression inside the square root as
\displaystyle 4-x^2=2^2-x^2
Therefore, \displaystyle I=\int \sqrt{2^2-x^2}\,dx
This is of the standard form
\displaystyle \int \sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\!\left(\frac{x}{a}\right)+C
Using the above formula with a=2, we get
\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}\sin^{-1}\!\left(\frac{x}{2}\right)+C
Simplifying,
\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\!\left(\frac{x}{2}\right)+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\!\left(\frac{x}{2}\right)+C}
Method 2: Using Trigonometric Substitution
In the given integral, \displaystyle I=\int \sqrt{4-x^2}\,dx
Put x=2\sin\theta
\displaystyle \Rightarrow dx=2\cos\theta\,d\theta
Also, \displaystyle \sqrt{4-x^2}=\sqrt{4-4\sin^2\theta}
\displaystyle =\sqrt{4(1-\sin^2\theta)}
\displaystyle =\sqrt{4\cos^2\theta}=2\cos\theta
Substituting in the integral, we get
\displaystyle I=\int (2\cos\theta)(2\cos\theta)\,d\theta
\displaystyle =4\int\cos^2\theta\,d\theta
Using the identity
\displaystyle \cos^2\theta=\frac{1+\cos2\theta}{2}
We get, \displaystyle I=4\int\frac{1+\cos2\theta}{2}\,d\theta
\displaystyle =2\int(1+\cos2\theta)\,d\theta
\displaystyle =2\left(\theta+\frac{\sin2\theta}{2}\right)+C
Thus, \displaystyle I=2\theta+\sin2\theta+C …(1)
Now consider, \displaystyle \sin2\theta=2\sin\theta\cos\theta
Since x=2\sin\theta,
i.e. \displaystyle \sin\theta=\frac{x}{2}
And using, \displaystyle \cos\theta=\sqrt{1-\sin^2\theta}
\displaystyle =\sqrt{1-\left(\frac{x}{2}\right)^2}
\displaystyle =\sqrt{\frac{4-x^2}{4}}
We get, \displaystyle \cos\theta =\frac{\sqrt{4-x^2}}{2}
Therefore,
\displaystyle \sin2\theta=2\cdot\frac{x}{2}\cdot\frac{\sqrt{4-x^2}}{2}=\frac{x\sqrt{4-x^2}}{2}
Also, \displaystyle \theta=\sin^{-1}\left(\frac{x}{2}\right)
Substituting these values in (1), we obtain
\displaystyle I=2\sin^{-1}\!\left(\frac{x}{2}\right)+\frac{x}{2}\sqrt{4-x^2}+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\!\left(\frac{x}{2}\right)+C}
Method 3: Using Integration by Parts
Let, \displaystyle I=\int\sqrt{4-x^2}\,dx
Since the integrand contains only one function, we write it as the product of \sqrt{4-x^2} and 1. We then use the Method of Integration by Parts.
So, considering \displaystyle I=\int\underbrace{\sqrt{4-x^2}}_{\mathrm{I}}\;\underbrace{1}_{\mathrm{II}}\,dx
Applying the Integration by Parts formula,
\displaystyle \int uv\,dx=u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx
We get, \displaystyle I=\sqrt{4-x^2}\int dx-\int\left(\frac{d}{dx}\left(\sqrt{4-x^2}\right)\int dx\right)dx
Since \displaystyle \int dx=x and by Chain Rule, \displaystyle \frac{d}{dx}\left(\sqrt{4-x^2}\right)=\frac{-2x}{2\sqrt{4-x^2}}=-\frac{x}{\sqrt{4-x^2}}, we get
\displaystyle I=x\sqrt{4-x^2}+\int\frac{x^2}{\sqrt{4-x^2}}\,dx
Now writing \displaystyle x^2=4-(4-x^2)
Therefore,
\displaystyle I=x\sqrt{4-x^2}+\int\frac{4-(4-x^2)}{\sqrt{4-x^2}}\,dx
Splitting the numerator gives,
\displaystyle I=x\sqrt{4-x^2}+4\int\frac{dx}{\sqrt{4-x^2}}-\int\frac{4-x^2}{\sqrt{4-x^2}}\,dx
\displaystyle =x\sqrt{4-x^2}+4\int\frac{dx}{\sqrt{4-x^2}}-\int\sqrt{4-x^2}\,dx
Since \displaystyle \int\sqrt{4-x^2}\,dx=I, we have
\displaystyle I=x\sqrt{4-x^2}+4\int\frac{dx}{\sqrt{4-x^2}}-I
Therefore,
\displaystyle 2I=x\sqrt{4-x^2}+4\int\frac{dx}{\sqrt{4-x^2}}
Using the standard result
\displaystyle \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\left(\frac{x}{a}\right)+C
with a=2, we get
\displaystyle \int\frac{dx}{\sqrt{4-x^2}}=\sin^{-1}\left(\frac{x}{2}\right)
Substituting above, we get
\displaystyle 2I=x\sqrt{4-x^2}+4\sin^{-1}\left(\frac{x}{2}\right)+C
Dividing both sides by 2, we get
\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\left(\frac{x}{2}\right)+C}
Conclusion: This example shows that the standard formulas of Exercise 7.7 are not just results to memorise. They can also be obtained using the trigonometric substitutions introduced in Exercise 7.2 or by applying Integration by Parts. However, once these formulas are known, using the standard formula is the quickest and most efficient method for solving NCERT questions.
Question 2: Integrals Ex. 7.7
2. Evaluate \displaystyle \int \sqrt{1-4x^2}\,dx.
Solution
Let, \displaystyle I=\int \sqrt{1-4x^2}\,dx
Since the expression inside the square root contains 2x,
Let, t=2x
Then \displaystyle dt=2\,dx\qquad\Rightarrow\qquad dx=\frac12\,dt
Substituting, we get
\displaystyle I=\frac12\int\sqrt{1-t^2}\,dt
This is of the standard form
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=1, we get
\displaystyle I=\frac12\left[\frac{t}{2}\sqrt{1-t^2}+\frac12\sin^{-1}(t)\right]+C
\displaystyle =\frac{t}{4}\sqrt{1-t^2}+\frac14\sin^{-1}(t)+C
Substituting back, t=2x,
\displaystyle I=\frac{2x}{4}\sqrt{1-4x^2}+\frac14\sin^{-1}(2x)+C
\displaystyle =\frac{x}{2}\sqrt{1-4x^2}+\frac14\sin^{-1}(2x)+C
Hence,
\boxed{\displaystyle I=\frac{x}{2}\sqrt{1-4x^2}+\frac14\sin^{-1}(2x)+C}
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Question 3: Completing the Square
3. Evaluate \displaystyle \int \sqrt{x^2+4x+6}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{x^2+4x+6}\,dx
To complete the square, consider the quadratic expression inside the square root.
\displaystyle x^2+4x+6
Now, the coefficient of x is 4.
Half of this coefficient is \displaystyle \frac{4}{2}=2
Adding and subtracting its square i.e. \displaystyle 2^2=4, we get
\displaystyle x^2+4x+6=x^2+4x+4-4+6
\displaystyle =x^2+4x+4+2
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
We get \displaystyle x^2+4x+6=(x+2)^2+2
\displaystyle =(x+2)^2+(\sqrt2)^2
Therefore,
\displaystyle I=\int\sqrt{(x+2)^2+(\sqrt2)^2}\,dx
Let t=x+2
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{t^2+(\sqrt2)^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=\sqrt2, we get
\displaystyle I=\frac{t}{2}\sqrt{t^2+2}+\frac{(\sqrt2)^2}{2}\log\left|t+\sqrt{t^2+2}\right|+C
\displaystyle =\frac{t}{2}\sqrt{t^2+2}+\log\left|t+\sqrt{t^2+2}\right|+C
Substituting back, t=x+2, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{x^2+4x+6}\,dx=\\[4pt] \displaystyle \frac{x+2}{2}\sqrt{x^2+4x+6}\\[4pt] \displaystyle +\log\left|x+2+\sqrt{x^2+4x+6}\right|+C \end{gathered}}
Question 4: Integrals 7.7
4. Evaluate \displaystyle \int \sqrt{x^2+4x+1}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{x^2+4x+1}\,dx
To complete the square, consider the quadratic expression inside the square root.
\displaystyle x^2+4x+1
Now, the coefficient of x is 4.
Half of this coefficient is \displaystyle \frac{4}{2}=2
Adding and subtracting its square i.e. \displaystyle 2^2=4, we get
\displaystyle x^2+4x+1=x^2+4x+4-4+1
\displaystyle =x^2+4x+4-3
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
We get \displaystyle x^2+4x+1=(x+2)^2-3
\displaystyle =(x+2)^2-(\sqrt3)^2
Therefore, \displaystyle I=\int\sqrt{(x+2)^2-(\sqrt3)^2}\,dx
Let t=x+2
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{t^2-(\sqrt3)^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=\sqrt3, we get
\displaystyle I=\frac{t}{2}\sqrt{t^2-3}-\frac{(\sqrt3)^2}{2}\log\left|t+\sqrt{t^2-3}\right|+C
\displaystyle =\frac{t}{2}\sqrt{t^2-3}-\frac32\log\left|t+\sqrt{t^2-3}\right|+C
Substituting back, t=x+2, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{x^2+4x+1}\,dx=\\[4pt] \displaystyle \frac{x+2}{2}\sqrt{x^2+4x+1}\\[4pt] \displaystyle -\frac32\log\left|x+2+\sqrt{x^2+4x+1}\right|+C \end{gathered}}
Question 5: Integrals 7.7
5. Evaluate \displaystyle \int \sqrt{1-4x-x^2}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{1-4x-x^2}\,dx
To complete the square, first take -1 common from the terms containing x.
\displaystyle 1-4x-x^2=-(x^2+4x)+1
Now, the coefficient of x is 4.
Half of this coefficient is \displaystyle \frac{4}{2}=2
Adding and subtracting its square i.e. \displaystyle 2^2=4 inside the bracket, we get
\displaystyle 1-4x-x^2=-\left(x^2+4x+4-4\right)+1
\displaystyle =-\left(x^2+4x+4\right)+4+1
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
We get \displaystyle 1-4x-x^2=-(x+2)^2+5
\displaystyle =(\sqrt5)^2-(x+2)^2
Therefore, \displaystyle I=\int\sqrt{(\sqrt5)^2-(x+2)^2}\,dx
Let t=x+2
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{(\sqrt5)^2-t^2}\,dt
This is of the type
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=\sqrt5, we get
\displaystyle I=\frac{t}{2}\sqrt{5-t^2}+\frac{(\sqrt5)^2}{2}\sin^{-1}\left(\frac{t}{\sqrt5}\right)+C
\displaystyle =\frac{t}{2}\sqrt{5-t^2}+\frac52\sin^{-1}\left(\frac{t}{\sqrt5}\right)+C
Substituting back, t=x+2, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{1-4x-x^2}\,dx=\\[4pt] \displaystyle \frac{x+2}{2}\sqrt{1-4x-x^2}\\[4pt] \displaystyle +\frac52\sin^{-1}\left(\frac{x+2}{\sqrt5}\right)+C \end{gathered}}
Important: Whenever a quadratic expression of the form ax^2+bx+c appears under a square root, your first goal should be to complete the square. This usually reduces the integral to one of the three standard forms involving \sqrt{a^2-x^2}, \sqrt{x^2+a^2} or \sqrt{x^2-a^2}, allowing you to apply the appropriate standard formula directly.
Question 6: Integrals 7.7
6. Evaluate \displaystyle \int \sqrt{x^2+4x-5}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{x^2+4x-5}\,dx
To complete the square, consider the quadratic expression inside the square root.
\displaystyle x^2+4x-5
Now, the coefficient of x is 4.
Half of this coefficient is \displaystyle \frac{4}{2}=2
Adding and subtracting its square i.e. \displaystyle 2^2=4, we get
\displaystyle x^2+4x-5=x^2+4x+4-4-5
\displaystyle =x^2+4x+4-9
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
We get \displaystyle x^2+4x-5=(x+2)^2-9
\displaystyle =(x+2)^2-3^2
Therefore, \displaystyle I=\int\sqrt{(x+2)^2-3^2}\,dx
Let t=x+2
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{t^2-3^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=3, we get
\displaystyle I=\frac{t}{2}\sqrt{t^2-9}-\frac{3^2}{2}\log\left|t+\sqrt{t^2-9}\right|+C
\displaystyle =\frac{t}{2}\sqrt{t^2-9}-\frac92\log\left|t+\sqrt{t^2-9}\right|+C
Substituting back, t=x+2, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{x^2+4x-5}\,dx=\\[4pt] \displaystyle \frac{x+2}{2}\sqrt{x^2+4x-5}\\[4pt] \displaystyle -\frac92\log\left|x+2+\sqrt{x^2+4x-5}\right|+C \end{gathered}}
Question 7: Exercise 7.7
7. Evaluate \displaystyle \int \sqrt{1+3x-x^2}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{1+3x-x^2}\,dx
To complete the square, first take -1 common from the terms containing x.
\displaystyle 1+3x-x^2=-(x^2-3x)+1
Now, the coefficient of x is -3.
Half of this coefficient is \displaystyle \frac{-3}{2}=-\frac32
Adding and subtracting its square i.e. \displaystyle \left(\frac32\right)^2=\frac94 inside the bracket,
We get, \displaystyle 1+3x-x^2=-\left(x^2-3x+\frac94-\frac94\right)+1
\displaystyle =-\left(x^2-3x+\frac94\right)+\frac94+1
\displaystyle =-\left(x^2-3x+\frac94\right)+\frac{13}{4}
Using the identity
\displaystyle a^2-2ab+b^2=(a-b)^2
We get \displaystyle 1+3x-x^2=-\left(x-\frac32\right)^2+\frac{13}{4}
\displaystyle =\left(\frac{\sqrt{13}}2\right)^2-\left(x-\frac32\right)^2
Therefore,
\displaystyle I=\int\sqrt{\left(\frac{\sqrt{13}}2\right)^2-\left(x-\frac32\right)^2}\,dx
Let t=x-\frac32
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{\left(\frac{\sqrt{13}}2\right)^2-t^2}\,dt
This is of the type
\displaystyle \int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C
Using the above formula with a=\dfrac{\sqrt{13}}2, we get
\displaystyle I=\frac{t}{2}\sqrt{a^2-t^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{t}{a}\right)+C
\displaystyle =\frac{t}{2}\sqrt{a^2-t^2}+\frac{13}{8}\sin^{-1}\left(\frac{2t}{\sqrt{13}}\right)+C
Substituting back, t=x-\frac32, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{1+3x-x^2}\,dx=\\[4pt] \displaystyle \frac{2x-3}{4}\sqrt{1+3x-x^2}\\[4pt] \displaystyle +\frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+C \end{gathered}}
Question 8: Standard Form of Integrals
8. Evaluate \displaystyle \int \sqrt{x^2+3x}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{x^2+3x}\,dx
To complete the square, consider the quadratic expression inside the square root.
\displaystyle x^2+3x
Now, the coefficient of x is 3.
Half of this coefficient is \displaystyle \frac{3}{2}
Adding and subtracting its square i.e. \displaystyle \left(\frac32\right)^2=\frac94, we get
\displaystyle x^2+3x=x^2+3x+\frac94-\frac94
\displaystyle =x^2+3x+\frac94-\frac94
Using the identity
\displaystyle a^2+2ab+b^2=(a+b)^2
we get
\displaystyle x^2+3x=\left(x+\frac32\right)^2-\frac94
\displaystyle =\left(x+\frac32\right)^2-\left(\frac32\right)^2
Therefore,
\displaystyle I=\int\sqrt{\left(x+\frac32\right)^2-\left(\frac32\right)^2}\,dx
Let t=x+\frac32
Then dt=dx
Substituting, we get
\displaystyle I=\int\sqrt{t^2-\left(\frac32\right)^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=\dfrac32, we get
\displaystyle I=\frac{t}{2}\sqrt{t^2-\frac94}-\frac{\left(\frac32\right)^2}{2}\log\left|t+\sqrt{t^2-\frac94}\right|+C
\displaystyle =\frac{t}{2}\sqrt{t^2-\frac94}-\frac98\log\left|t+\sqrt{t^2-\frac94}\right|+C
Substituting back, t=x+\frac32, therefore
\boxed{\begin{gathered}\displaystyle \int\sqrt{x^2+3x}\,dx=\\[4pt] \displaystyle \frac{2x+3}{4}\sqrt{x^2+3x}\\[4pt] \displaystyle -\frac98\log\left|x+\frac32+\sqrt{x^2+3x}\right|+C \end{gathered}}
By now, you have probably noticed that the key to most questions in this exercise is recognizing the pattern hidden inside the integrand. Once the expression is reduced to a standard form, the integration becomes much simpler. The more such questions you solve, the faster you’ll become at spotting the right approach.
If you prefer learning through videos, many of these concepts and questions are also explained on my YouTube Channel @MathsBetter. Watching the solutions alongside practice can help you build speed and confidence.
Now, let’s proceed to the next question.
Question 9: Integrals 7.7
9. Evaluate \displaystyle \int \sqrt{1+\frac{x^2}{9}}\,dx.
Solution
Let, \displaystyle I=\int\sqrt{1+\frac{x^2}{9}}\,dx
Since the expression inside the square root contains \dfrac{x}{3},
Let t=\frac{x}{3}
Then \displaystyle dt=\frac13\,dx\qquad\Rightarrow\qquad dx=3\,dt
Substituting, we get
\displaystyle I=3\int\sqrt{1+t^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C
Using the above formula with a=1, we get
\displaystyle I=3\left[\frac{t}{2}\sqrt{1+t^2}+\frac12\log\left|t+\sqrt{1+t^2}\right|\right]+C
\displaystyle =\frac{3t}{2}\sqrt{1+t^2}+\frac32\log\left|t+\sqrt{1+t^2}\right|+C
Substituting back, t=\dfrac{x}{3}, we get
\displaystyle I=\frac{x}{2}\sqrt{1+\frac{x^2}{9}}+\frac32\log\left|\frac{x}{3}+\sqrt{1+\frac{x^2}{9}}\right|+C
Now,
\displaystyle \sqrt{1+\frac{x^2}{9}}=\sqrt{\frac{x^2+9}{9}}=\frac{\sqrt{x^2+9}}{3}
Therefore, \displaystyle I=\frac{x}{2}\cdot\frac{\sqrt{x^2+9}}{3}+\frac32\log\left|\frac{x+\sqrt{x^2+9}}{3}\right|+C
\displaystyle =\frac{x}{6}\sqrt{x^2+9}+\frac32\left(\log\left|x+\sqrt{x^2+9}\right|-\log3\right)+C
\displaystyle =\frac{x}{6}\sqrt{x^2+9}+\frac32\log\left|x+\sqrt{x^2+9}\right|-\frac32\log3+C
Hence
\boxed{\displaystyle I=\frac{x}{6}\sqrt{x^2+9}+\frac32\log\left|x+\sqrt{x^2+9}\right|+C'}
where \displaystyle C'=C-\frac32\log3
Question 10: Integrals 7.7 – MCQ
10. Choose the correct answer.
\displaystyle \int \sqrt{1+x^2}\,dx is equal to
- (A) \displaystyle \frac{x}{2}\sqrt{1+x^2}+\frac12\log\left|x+\sqrt{1+x^2}\right|+C
- (B) \displaystyle \frac23(1+x^2)^{3/2}+C
- (C) \displaystyle \frac23x(1+x^2)^{3/2}+C
- (D) \displaystyle \frac{x^2}{2}\sqrt{1+x^2}+\frac12x^2\log\left|x+\sqrt{1+x^2}\right|+C
Solution
From Question 9 (or using the standard formula), we know that
\displaystyle \int\sqrt{x^2+a^2}\,dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log\left|x+\sqrt{x^2+a^2}\right|+C
Here, a=1. Therefore,
\displaystyle \int\sqrt{1+x^2}\,dx=\frac{x}{2}\sqrt{1+x^2}+\frac12\log\left|x+\sqrt{1+x^2}\right|+C
✅️ Hence, the correct answer is (A).
Question 11: Integrals 7.7 – MCQ
11. Choose the correct answer.
\displaystyle \int \sqrt{x^2-8x+7}\,dx is equal to
- (A) \displaystyle \frac12(x-4)\sqrt{x^2-8x+7}+9\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
- (B) \displaystyle \frac12(x+4)\sqrt{x^2-8x+7}+9\log\left|x+4+\sqrt{x^2-8x+7}\right|+C
- (C) \displaystyle \frac12(x-4)\sqrt{x^2-8x+7}-3\sqrt2\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
- (D) \displaystyle \frac12(x-4)\sqrt{x^2-8x+7}-\frac92\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
Solution
Complete the square:
\displaystyle x^2-8x+7=x^2-8x+16-16+7
\displaystyle =(x-4)^2-9
Therefore,
\displaystyle \int\sqrt{x^2-8x+7}\,dx=\int\sqrt{(x-4)^2-3^2}\,dx
Let t=x-4
Then dt=dx
Substituting, we get
\displaystyle \int\sqrt{t^2-3^2}\,dt
This is of the type
\displaystyle \int\sqrt{x^2-a^2}\,dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log\left|x+\sqrt{x^2-a^2}\right|+C
Using the above formula with a=3, we get
\displaystyle \int\sqrt{t^2-9}\,dt=\frac{t}{2}\sqrt{t^2-9}-\frac92\log\left|t+\sqrt{t^2-9}\right|+C
Substituting back, t=x-4, we get
\displaystyle I=\frac12(x-4)\sqrt{x^2-8x+7}-\frac92\log\left|x-4+\sqrt{x^2-8x+7}\right|+C
✅️ Hence, the correct answer is (D).
Common Mistakes to Avoid
- Not identifying the correct standard form: Before integrating, rewrite the given integral so that it matches one of the standard forms involving \sqrt{a^2-x^2}, \sqrt{x^2+a^2} or \sqrt{x^2-a^2}.
- Skipping the completion of the square: Whenever the expression inside the square root is a quadratic, complete the square first before applying any standard formula.
- Making errors while completing the square: Always divide the coefficient of x by 2, square it, and then add as well as subtract the same quantity. Avoid jumping directly to the final result.
- Using the wrong standard formula: The three formulas look similar but give different answers involving \sin^{-1}x or logarithms. Check carefully whether the integrand is of the form \sqrt{a^2-x^2}, \sqrt{x^2+a^2} or \sqrt{x^2-a^2}.
- Forgetting the linear substitution: After completing the square, use a suitable substitution such as t=x+h to reduce the integral to a standard form.
- Not substituting back in terms of x: After integrating in terms of t, always substitute back to obtain the final answer in the original variable.
- Making mistakes while simplifying logarithmic answers: Simplify radicals and logarithmic expressions carefully. Any constant obtained from logarithmic simplification can be absorbed into the constant of integration.
- Forgetting the constant of integration: Every indefinite integral must end with C (or an equivalent constant such as C').
Continue Learning
After completing Exercise 7.7 of Integrals, you should now be comfortable with evaluating integrals involving the square root of quadratic expressions. In this exercise, you learned how to reduce a given integral to one of the three standard forms by using a suitable substitution or by completing the square. You also saw that these standard formulas can be derived using trigonometric substitution or Integration by Parts.
To strengthen your understanding further, make sure that you revise:
- The three standard integral formulas involving \sqrt{a^2-x^2}, \sqrt{x^2+a^2} and \sqrt{x^2-a^2}.
- Reducing a given integral to a standard form using a suitable linear substitution such as t=ax+b.
- Completing the square in quadratic expressions of the form ax^2+bx+c.
- Identifying whether the integral reduces to the form \sqrt{a^2-x^2}, \sqrt{x^2+a^2} or \sqrt{x^2-a^2}.
- Applying the correct standard formula involving \sin^{-1}x or logarithmic functions.
- Simplifying the final answer by substituting back the original variable and using logarithmic properties wherever required.
- Remembering that the same standard formulas can also be obtained using trigonometric substitution or Integration by Parts.
- Writing the final answer in its simplest form along with the constant of integration C.
Explore More
Integrals involving square roots of quadratic expressions become much easier once you learn to recognise the underlying pattern. As you practise more questions, focus on reducing the given integral to one of the three standard forms by using a suitable substitution or completing the square carefully. Then apply the appropriate standard formula, simplify the final answer wherever possible and include the constant of integration C. With regular practice, these techniques will become natural and help you solve even complicated integrals with confidence.
All the best and keep learning 👍
