Integrals: Chapter 7 Links
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Integrals 7.9 extends the concept of definite integrals by introducing the method of substitution. While the technique of substitution is similar to that used in indefinite integrals, the limits of integration must also be changed according to the new variable. This allows us to evaluate definite integrals directly without converting back to the original variable.
\displaystyle \text{If }x=\phi(t),\text{ then }\int_a^b f(x)\,dx=\int_{\alpha}^{\beta}f(\phi(t))\,\phi'(t)\,dt,\quad\text{where }\alpha,\beta\text{ are the new limits.}
Changing the limits after substitution makes the evaluation simpler and avoids the extra step of rewriting the answer in terms of the original variable. This method is especially useful for integrals involving trigonometric, exponential and algebraic substitutions.
In this exercise, we will learn:
- how to evaluate definite integrals using substitution.
- how to change the limits of integration after introducing a new variable.
- how the method of substitution simplifies the evaluation of definite integrals.
- when it is unnecessary to substitute back to the original variable.
- how to solve all NCERT Exercise 7.9 questions step by step using substitution and change of limits.
Key Concepts
1. Integration by Substitution
Integration by substitution is used when the integrand contains a function along with its derivative. We substitute a suitable variable to simplify the integral.
If u=g(x), then
\displaystyle \int f(g(x))\,g'(x)\,dx=\int f(u)\,du
For definite integrals, the limits of integration are also changed according to the new variable.
2. Changing the Limits of Integration
After substituting a new variable, the limits must be expressed in terms of that variable. This eliminates the need to substitute back to the original variable.
If u=g(x), then
\displaystyle \int_a^b f(g(x))\,g'(x)\,dx=\int_{g(a)}^{g(b)}f(u)\,du
Always evaluate the new limits before solving the integral.
3. Steps for Integration by Substitution
- Choose a suitable substitution u=g(x).
- Differentiate to find du=g'(x)\,dx.
- Change the limits of integration to the new variable.
- Evaluate the integral using the new limits.
- Do not substitute back to the original variable.
Tip: While solving a definite integral by substitution, always change the limits of integration to the new variable before evaluating the integral. This avoids substituting back to the original variable, making the solution shorter, simpler and less prone to errors.
Let us now solve all the NCERT questions step by step in Integrals 7.9.
Question 1: Definite Integrals 7.9
1. Evaluate \displaystyle \int_{0}^{1}\frac{x}{x^2+1}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{1}\frac{x}{x^2+1}\,dx
Using the substitution
u=x^2+1
we get
\displaystyle \frac{du}{dx}=2x\quad\Rightarrow\quad du=2x\,dx
or \displaystyle x\,dx=\frac{du}{2}
Now, change the limits of integration:
When x=0, u=0^2+1=1
When x=1, u=1^2+1=2
Therefore, the given integral reduces to
\displaystyle I=\frac12\int_{1}^{2}\frac{du}{u}
\displaystyle =\frac12\left[\log|u|\right]_{1}^{2}
Substituting the limits, we get
\displaystyle I=\frac12\left(\log2-\log1\right)
Since \log1=0,
\displaystyle \therefore I=\frac12\log2
Hence,
\boxed{\displaystyle \int_{0}^{1}\frac{x}{x^2+1}\,dx=\frac12\log2}
Question 2: Definite Integrals 7.9
2. Evaluate \displaystyle \int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi
Since \cos^5\phi=\cos^4\phi\cos\phi=(1-\sin^2\phi)^2\cos\phi, we have
\displaystyle I=\int_{0}^{\pi/2}\sqrt{\sin\phi}\,(1-\sin^2\phi)^2\cos\phi\,d\phi
Using the substitution
u=\sin\phi
we get
\displaystyle \frac{du}{d\phi}=\cos\phi\quad\Rightarrow\quad du=\cos\phi\,d\phi
Now, change the limits of integration:
When \phi=0, u=\sin0=0
When \phi=\dfrac{\pi}{2}, u=\sin\dfrac{\pi}{2}=1
Therefore, the given integral reduces to
\displaystyle I=\int_{0}^{1}u^{1/2}(1-u^2)^2\,du
\displaystyle =\int_{0}^{1}\left(u^{1/2}-2u^{5/2}+u^{9/2}\right)du
Integrating the respective terms, we get
\displaystyle I=\left[\frac{2}{3}u^{3/2}-\frac{4}{7}u^{7/2}+\frac{2}{11}u^{11/2}\right]_0^1
Substituting the limits,
We get, \displaystyle I=\left(\frac23-\frac47+\frac2{11}\right)-0
\displaystyle =\frac{154-132+42}{231}=\frac{64}{231}
\displaystyle \therefore I=\frac{64}{231}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi=\frac{64}{231}}
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In the same way, Integrals 7.9 for Class 12 Maths is designed to strengthen your concepts and improve step-by-step problem-solving skills. Now, let’s proceed to the next question.
Question 3: Integrals 7.9
3. Evaluate \displaystyle \int_{0}^{1}\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx.
Solution
Let, \displaystyle I=\int_{0}^{1}\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx
Using the substitution
\displaystyle x=\tan\theta we get, \displaystyle dx=\sec^2\theta\,d\theta
Also,
\displaystyle \frac{2x}{1+x^2}=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\tan\theta}{\sec^2\theta}=2\sin\theta\cos\theta=\sin2\theta
Now, change the limits of integration:
When x=0, i.e. \tan\theta=0, we get \theta=\tan^{-1}(0)=0
When x=1, i.e. \tan\theta=1, we get \theta=\tan^{-1}(1)=\dfrac{\pi}{4}
Since 0\le\theta\le\dfrac{\pi}{4}, therefore 0\le2\theta\le\dfrac{\pi}{2}. Hence, \sin^{-1}(\sin2\theta)=2\theta
Therefore, the given integral reduces to
\displaystyle I=\int_{0}^{\pi/4}2\theta\sec^2\theta\,d\theta
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=2\theta\int\sec^2\theta\,d\theta-\int\left(\frac{d}{d\theta}(2\theta)\int\sec^2\theta\,d\theta\right)d\theta
Since \displaystyle \int\sec^2\theta\,d\theta=\tan\theta and \displaystyle \frac{d}{d\theta}(2\theta)=2, we get
\displaystyle I=2\theta\tan\theta-2\int\tan\theta\,d\theta
Using \displaystyle \int\tan x\,dx=\log|\sec x|=-\log|\cos x|, we obtain
\displaystyle I=2\theta\tan\theta+2\log|\cos\theta|
Applying the limits,
We get, \displaystyle I=\left[2\theta\tan\theta+2\log|\cos\theta|\right]_{0}^{\pi/4}
\displaystyle =\left(2\cdot\frac{\pi}{4}\cdot\tan\frac{\pi}{4}+2\log\left|\cos\frac{\pi}{4}\right|\right)-\left(2\cdot0\cdot\tan0+2\log|\cos0|\right)
\displaystyle =\frac{\pi}{2}+2\left(\log\frac1{\sqrt2}-\log1\right)
Using \log1=0 and logarithms properties,
\displaystyle =\frac{\pi}{2}+2\log\frac1{\sqrt2}
\displaystyle =\frac{\pi}{2}-\log2\qquad\left(\because \log\frac1{\sqrt2}=-\frac12\log2\right)
So, \displaystyle I=\frac{\pi}{2}-\log2
Hence,
\boxed{\displaystyle \int_{0}^{1}\sin^{-1}\!\left(\frac{2x}{1+x^2}\right)\,dx=\frac{\pi}{2}-\log2}
Question 4: Integrals Exercise 7.9
4. Evaluate \displaystyle \int_{0}^{2}x\sqrt{x+2}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{2}x\sqrt{x+2}\,dx
Using the substitution
\displaystyle x+2=t^2
we get
\displaystyle x=t^2-2
Differentiating both sides with respect to t, we get
\displaystyle \frac{dx}{dt}=2t\quad\Rightarrow\quad dx=2t\,dt
Also,
\displaystyle \sqrt{x+2}=\sqrt{t^2}=t
Now, change the limits of integration:
When x=0, i.e. t^2=2, we get t=\sqrt2
When x=2, i.e. t^2=4, we get t=2
Therefore, the given integral reduces to
\displaystyle I=\int_{\sqrt2}^{2}(t^2-2)\cdot t\cdot2t\,dt
\displaystyle =2\int_{\sqrt2}^{2}(t^4-2t^2)\,dt
Integrating
\displaystyle =2\left[\frac{t^5}{5}-\frac{2t^3}{3}\right]_{\sqrt2}^{2}
Applying the limits,
We get, \displaystyle I=2\left[\left(\frac{2^5}{5}-\frac{2\cdot2^3}{3}\right)-\left(\frac{(\sqrt2)^5}{5}-\frac{2(\sqrt2)^3}{3}\right)\right]
\displaystyle =2\left[\left(\frac{32}{5}-\frac{16}{3}\right)-\left(\frac{4\sqrt2}{5}-\frac{4\sqrt2}{3}\right)\right]
or \displaystyle I=2\left[\frac{16}{15}+\frac{8\sqrt2}{15}\right]
\displaystyle =\frac{32+16\sqrt2}{15}=\frac{16\sqrt2}{15}(\sqrt2+1)
\displaystyle \therefore I=\frac{16\sqrt2}{15}(\sqrt2+1)
Hence,
\boxed{\displaystyle \int_{0}^{2}x\sqrt{x+2}\,dx=\frac{16\sqrt2}{15}(\sqrt2+1)}
Question 5: Definite Integrals 7.9
5. Evaluate \displaystyle \int_{0}^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx.
Solution
Let, \displaystyle I=\int_{0}^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx
Using the substitution
\displaystyle t=\cos x
we get
\displaystyle \frac{dt}{dx}=-\sin x\quad\Rightarrow\quad dt=-\sin x\,dx
or, \displaystyle \sin x\,dx=-dt
Now, change the limits of integration:
When x=0, i.e. t=\cos0, we get t=1
When x=\dfrac{\pi}{2}, i.e. t=\cos\dfrac{\pi}{2}, we get t=0
Therefore, the given integral reduces to
\displaystyle I=-\int_{1}^{0}\frac{dt}{1+t^2}
Note: The negative sign can also be eliminated by reversing the limits of integration. This is a useful property of definite integrals, which will be studied in Exercise 7.10. Here, we evaluate the integral directly without using that property.
\displaystyle =-\left[\tan^{-1}t\right]_{1}^{0}
Applying the limits,
We get, \displaystyle I=-\left(\tan^{-1}0-\tan^{-1}1\right)
\displaystyle =-\left(0-\frac{\pi}{4}\right)=\frac{\pi}{4}
\displaystyle \therefore I=\frac{\pi}{4}
Hence,
\boxed{\displaystyle \int_{0}^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx=\frac{\pi}{4}}
Interesting Fact: One of the biggest advantages of definite integrals is that after making a suitable substitution, the limits of integration can also be changed. This eliminates the need to substitute the original variable back into the integral, making the solution shorter and reducing the chances of mistakes.
Question 6: Integrals 7.9
6. Evaluate \displaystyle \int_{0}^{2}\frac{dx}{x+4-x^2}.
Solution
Let, \displaystyle I=\int_{0}^{2}\frac{dx}{x+4-x^2}
First, simplify the denominator by completing the square.
\displaystyle x+4-x^2=-(x^2-x-4)
Adding and subtracting \frac14.
\displaystyle =-\left(x^2-x+\frac14-\frac14-4\right)
\displaystyle =-\left[\left(x-\frac12\right)^2-\frac{17}{4}\right]
or
\displaystyle =\frac{17}{4}-\left(x-\frac12\right)^2
\displaystyle =\frac14\left[17-(2x-1)^2\right]
Therefore,
\displaystyle I=4\int_{0}^{2}\frac{dx}{17-(2x-1)^2}
Using the substitution
\displaystyle u=2x-1
we get
\displaystyle \frac{du}{dx}=2\quad\Rightarrow\quad du=2\,dx
or \displaystyle dx=\frac{du}{2}
Now, change the limits of integration:
When x=0, i.e. u=2(0)-1, we get u=-1
When x=2, i.e. u=2(2)-1, we get u=3
Therefore, the given integral reduces to
\displaystyle I=2\int_{-1}^{3}\frac{du}{17-u^2}
Using the standard result
\displaystyle \int\frac{dx}{a^2-x^2}=\frac1{2a}\log\left|\frac{a+x}{a-x}\right|+C
where a=\sqrt{17}, we get
\displaystyle I=\frac1{\sqrt{17}}\left[\log\left|\frac{\sqrt{17}+u}{\sqrt{17}-u}\right|\right]_{-1}^{3}
Applying the limits,
We get, \displaystyle I=\frac1{\sqrt{17}}\left[\log\left(\frac{\sqrt{17}+3}{\sqrt{17}-3}\right)-\log\left(\frac{\sqrt{17}-1}{\sqrt{17}+1}\right)\right]
\displaystyle =\frac1{\sqrt{17}}\log\left[\frac{\sqrt{17}+3}{\sqrt{17}-3}\times\frac{\sqrt{17}+1}{\sqrt{17}-1}\right]\qquad(\because \log a-\log b=\log\frac ab)
\displaystyle =\frac1{\sqrt{17}}\log\left[\frac{(\sqrt{17}+3)(\sqrt{17}+1)}{(\sqrt{17}-3)(\sqrt{17}-1)}\right]
On simplification.
\displaystyle =\frac1{\sqrt{17}}\log\left(\frac{20+4\sqrt{17}}{20-4\sqrt{17}}\right)
\displaystyle =\frac1{\sqrt{17}}\log\left(\frac{5+\sqrt{17}}{5-\sqrt{17}}\right)
Rationalizing the denominator inside the logarithm,
We get, \displaystyle I=\frac1{\sqrt{17}}\log\left(\frac{(5+\sqrt{17})^2}{25-17}\right)
\displaystyle =\frac1{\sqrt{17}}\log\left(\frac{25+17+10\sqrt{17}}8\right)
or \displaystyle I=\frac1{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}4\right)
\displaystyle \therefore I=\frac1{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}4\right)
Hence,
\boxed{\displaystyle \int_{0}^{2}\frac{dx}{x+4-x^2}=\frac1{\sqrt{17}}\log\left(\frac{21+5\sqrt{17}}4\right)}
Interesting Fact: Unlike an indefinite integral, which represents a family of functions, a definite integral always evaluates to a fixed numerical value. This is why the constant of integration C is not included in its final answer.
Question 7: Exercise 7.9
7. Evaluate \displaystyle \int_{-1}^{1}\frac{dx}{x^2+2x+5}.
Solution
Let, \displaystyle I=\int_{-1}^{1}\frac{dx}{x^2+2x+5}
First, rewrite the denominator, as a complete square
\displaystyle x^2+2x+5=(x+1)^2+4
Therefore,
\displaystyle I=\int_{-1}^{1}\frac{dx}{(x+1)^2+4}
Using the substitution
\displaystyle u=x+1
we get
\displaystyle \frac{du}{dx}=1\quad\Rightarrow\quad du=dx
Now, change the limits of integration:
When x=-1, i.e. u=-1+1, we get u=0
When x=1, i.e. u=1+1, we get u=2
Therefore, the given integral reduces to
\displaystyle I=\int_{0}^{2}\frac{du}{u^2+4}
Using the standard result
\displaystyle \int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\!\left(\frac{x}{a}\right)+C
where a=2, we get
\displaystyle I=\frac12\left[\tan^{-1}\!\left(\frac{u}{2}\right)\right]_{0}^{2}
Applying the limits,
We get, \displaystyle I=\frac12\left(\tan^{-1}1-\tan^{-1}0\right)
\displaystyle =\frac12\left(\frac{\pi}{4}\right)=\frac{\pi}{8}
\displaystyle \therefore I=\frac{\pi}{8}
Hence,
\boxed{\displaystyle \int_{-1}^{1}\frac{dx}{x^2+2x+5}=\frac{\pi}{8}}
Question 8: Integrals Ex. 7.9
8. Evaluate \displaystyle \int_{1}^{2}\left(\frac1x-\frac1{2x^2}\right)e^{2x}\,dx.
Solution (Method 1: Using Integration by Parts)
Let, \displaystyle I=\int_{1}^{2}\left(\frac1x-\frac1{2x^2}\right)e^{2x}\,dx
Separating the given integral, we get
\displaystyle I=\int_{1}^{2}\frac{e^{2x}}x\,dx-\frac12\int_{1}^{2}\frac{e^{2x}}{x^2}\,dx
Applying Integration by Parts to the first integral.
Using the ILATE Rule, we choose \displaystyle \frac1x as the first function and e^{2x} as the second function.
\displaystyle \int_{1}^{2}\underbrace{\frac1x}_{\text{I}}\underbrace{e^{2x}}_{\text{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 \int_{1}^{2}\frac{e^{2x}}x\,dx=\left[\frac1x\int e^{2x}\,dx\right]_{1}^{2}-\int_{1}^{2}\left(\frac{d}{dx}\frac1x\int e^{2x}\,dx\right)dx
Since \displaystyle \int e^{2x}\,dx=\frac{e^{2x}}2 and \displaystyle \frac{d}{dx}\left(\frac1x\right)=-\frac1{x^2}, we get
\displaystyle \int_{1}^{2}\frac{e^{2x}}x\,dx=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}+\frac12\int_{1}^{2}\frac{e^{2x}}{x^2}\,dx
Substituting this in the given expression for I, we get
\displaystyle I=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}+\frac12\int_{1}^{2}\frac{e^{2x}}{x^2}\,dx-\frac12\int_{1}^{2}\frac{e^{2x}}{x^2}\,dx
The last two integrals cancel each other.
Therefore,
\displaystyle I=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}
Applying the limits, we get
\displaystyle I=\frac{e^4}{4}-\frac{e^2}{2}
\displaystyle \therefore I=\frac{e^2}{4}(e^2-2)
Hence,
\boxed{\displaystyle \int_{1}^{2}\left(\frac1x-\frac1{2x^2}\right)e^{2x}\,dx=\frac{e^2}{4}(e^2-2)}
Method 2: Using Observation
Taking \displaystyle \frac12 common, we get
\displaystyle I=\frac12\int_{1}^{2}\left(\frac2x-\frac1{x^2}\right)e^{2x}\,dx
Let \displaystyle f(x)=\frac1x
Then \displaystyle f'(x)=-\frac1{x^2}
Hence,
\displaystyle I=\frac12\int_{1}^{2}\left(2f(x)+f'(x)\right)e^{2x}\,dx
Since
\displaystyle \frac{d}{dx}\left(e^{2x}f(x)\right)=e^{2x}\left(2f(x)+f'(x)\right)
\displaystyle \Rightarrow e^{2x}f(x)=\int e^{2x}\left(2f(x)+f'(x)\right)\,dx
Therefore,
\displaystyle I=\frac12\left[\frac{e^{2x}}x\right]_{1}^{2}
\displaystyle =\frac12\left(\frac{e^4}{2}-e^2\right)=\frac{e^2}{4}(e^2-2)
Hence,
\boxed{\displaystyle \int_{1}^{2}\left(\frac1x-\frac1{2x^2}\right)e^{2x}\,dx=\frac{e^2}{4}(e^2-2)}
Congratulations on completing NCERT Class 12 Maths Part 1! We have covered all the chapters and exercises with detailed explanations and step-by-step solutions. The journey continues with Part 2, where I’ll keep providing easy-to-follow solutions and concept-based explanations. Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, to help you learn and revise more effectively.
Now, let’s proceed to the two important MCQs of Definite Integrals 7.9.
Question 9: Integrals 7.9 – MCQ
9. Choose the correct answer.
The value of the integral \displaystyle \int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx is
- (A) 6
- (B) 0
- (C) 3
- (D) 4
Solution
Let, \displaystyle I=\int_{1/3}^{1}\frac{(x-x^3)^{1/3}}{x^4}\,dx
Taking x^3 common from x-x^3,
We get, \displaystyle I=\int_{1/3}^{1}\frac{\left[x^3\left(\frac1{x^2}-1\right)\right]^{1/3}}{x^4}\,dx
\displaystyle =\int_{1/3}^{1}\frac{x\left(\frac1{x^2}-1\right)^{1/3}}{x^4}\,dx
\displaystyle =\int_{1/3}^{1}\frac{\left(\frac1{x^2}-1\right)^{1/3}}{x^3}\,dx
Using the substitution
\displaystyle u=\frac1{x^2}-1
we get
\displaystyle \frac{du}{dx}=-\frac2{x^3}\quad\Rightarrow\quad du=-\frac2{x^3}\,dx
or \displaystyle \frac{dx}{x^3}=-\frac{du}{2}
Now, change the limits of integration:
When x=\dfrac13, i.e. \displaystyle u=\frac{1}{(1/3)^2}-1, we get u=8
When x=1, i.e. \displaystyle u=\frac11-1, we get u=0
Therefore, the given integral reduces to
\displaystyle I=-\frac12\int_8^0u^{1/3}\,du
\displaystyle =-\frac12\left[\frac34u^{4/3}\right]_8^0
Applying the limits,
We get, \displaystyle I=-\frac12\left(0-\frac34\times8^{4/3}\right)
\displaystyle =\frac38\times16=6
\displaystyle \boxed{\therefore I=6}
✅️ Hence, the correct option is (A).
Question 10: Definite Integrals 7.9 – MCQ
10. Choose the correct answer.
If \displaystyle f(x)=\int_{0}^{x}t\sin t\,dt, then f'(x) is
- (A) \cos x+x\sin x
- (B) x\sin x
- (C) x\cos x
- (D) \sin x+x\cos x
Solution
Given,
\displaystyle f(x)=\int_{0}^{x}t\sin t\,dt
By the First Fundamental Theorem of Calculus,
\displaystyle \frac{d}{dx}\left(\int_{a}^{x}F(t)\,dt\right)=F(x)
where F(t) is continuous in the given interval.
Here,
\displaystyle F(t)=t\sin t
Therefore,
\boxed {\displaystyle f'(x)=x\sin x}
✅️ Hence, the correct option is (B).
Alternative Method (Using Integration by Parts)
Given,
\displaystyle f(x)=\int_{0}^{x}t\sin t\,dt
Using Integration by Parts, we have
Using the ILATE Rule, we choose t as the first function and \sin t as the second function.
\displaystyle \int\underbrace{t}_{\text{I}}\underbrace{\sin t}_{\text{II}}\,dt
Applying the Integration by Parts formula,
\displaystyle \int uv\,dt=u\int v\,dt-\int\left(\frac{du}{dt}\int v\,dt\right)dt
We get
\displaystyle \int t\sin t\,dt=t(-\cos t)-\int(-\cos t)\,dt
\displaystyle =-t\cos t+\sin t
Therefore,
\displaystyle f(x)=\left[-t\cos t+\sin t\right]_0^x
Applying the limits, we get
\displaystyle f(x)=-x\cos x+\sin x
Differentiating both sides with respect to x using product rule,
\displaystyle f'(x)=-(\cos x-x\sin x)+\cos x
\boxed {\displaystyle \Rightarrow f'(x)=x\sin x}
✅️ Hence, the correct option is (B).
Common Mistakes to Avoid
- Not choosing a suitable substitution: Select a substitution that simplifies the integrand and whose derivative (or a constant multiple of it) is present in the integral.
- Forgetting to change the limits of integration: After substituting a new variable, evaluate the new lower and upper limits using the substitution before proceeding further.
- Making mistakes while changing the limits: Substitute the original limits carefully in the substitution equation. Do not assume the new limits without showing the intermediate step.
- Mixing variables: Once the substitution is made and the limits are changed, complete the integration entirely in the new variable. Avoid switching back to x unnecessarily.
- Using an incorrect standard integral: Identify the correct standard form after substitution, such as \displaystyle \int\frac{dx}{a^2-x^2}, \displaystyle \int\frac{dx}{x^2+a^2} or \displaystyle \int\frac{dx}{x}, before applying the corresponding formula.
- Ignoring algebraic simplification: Before making a substitution, simplify the integrand by factorising, completing the square or using algebraic identities wherever possible.
- Making errors while applying the new limits: Evaluate the anti-derivative carefully at both limits, especially when logarithmic, inverse trigonometric or exponential functions are involved.
- Using incorrect logarithmic properties: While simplifying the final answer, remember that \displaystyle \log a-\log b=\log\left(\frac ab\right). Apply logarithmic laws carefully.
- Using properties of definite integrals prematurely: In this exercise, evaluate the integral directly after changing the variable and the limits. The standard properties of definite integrals, such as reversing the limits, will be studied in the next exercise.
- Not writing the final answer in its simplest form: Simplify the result using logarithmic properties, algebraic simplification or factorisation wherever possible before writing the final answer.
Continue Learning
After completing Integrals 7.9, you should now be familiar with evaluating definite integrals using substitution by changing both the variable and the limits of integration. This method allows the entire integration to be carried out in the new variable, eliminating the need to substitute back to the original variable.
To strengthen your understanding further, make sure that you revise:
- Choosing a suitable substitution that simplifies the given integral.
- Finding the differential correctly and replacing dx in terms of the new variable.
- Changing the lower and upper limits of integration after substitution.
- Completing the integration entirely in the new variable without substituting back to x.
- Recognising standard integral forms after substitution, such as \displaystyle \int\frac{dx}{a^2-x^2}, \displaystyle \int\frac{dx}{x^2+a^2} and \displaystyle \int\frac{dx}{x}.
- Simplifying the integrand first by factorisation, completing the square or algebraic manipulation wherever required.
- Applying logarithmic, exponential and inverse trigonometric formulas correctly while evaluating the final answer.
- Using Integration by Parts or previously learned standard integrals whenever substitution alone is not sufficient.
- Evaluating the final expression carefully after applying the new limits of integration.
- Writing the final answer in its simplest form after the required algebraic, logarithmic or trigonometric simplifications.
Explore More
Definite integrals become much easier with regular practice. While solving questions in Integrals 7.9, first simplify the integrand wherever possible, then choose a suitable substitution, change the limits of integration accordingly, and complete the integration entirely in the new variable. Finally, apply the new limits carefully and simplify the result using the appropriate algebraic, logarithmic or trigonometric properties. With consistent practice, you’ll be able to identify suitable substitutions quickly and evaluate definite integrals accurately and confidently.
All the best and keep learning 👍
