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Definite Integration By Substitution

Learn substitution rule and method for evaluate definite integrals. Practice integration by substitution problems & Symmetric Functions & Properties of Definite Integrals.

Evaluating Definite Integrals by Substitution

Method 1   Use fundamental theorem of calculus after evaluating the indefinite integral.

\(\int\limits_a^bf'(x)\;dx=f(b)-f(a)\),  where

 \(\int\,f'(x)\;dx=f(x)+c\)

e.g. 

Consider , \(\int\limits_0^{\sqrt{ ({\pi}/{2}}) }x\,sin\;(x^2)\;dx\)

Put \(x^2=t \)

\(\Rightarrow\,2x\;dx=dt\)   

For upper limit when \(x = \sqrt{\dfrac{\pi}{2}}\) then  \(t=x^2=\dfrac{\pi}{2}\)

\(\therefore\; \int\limits_0^{({\pi}/{2})} sin\;t\;(x\;dx)\)

\(=\int\limits_{{0}}^{({\pi}/{2})} sin\;t\;\dfrac{dt}{2}\)

\(=\bigg[\dfrac {1}{2}×(-cos\;t) \bigg]_{{0}}^{({\pi}/{2})}\)  

\(=-\dfrac {1}{2}\bigg[\,cos\,\dfrac {\pi}{2}-{cos\,0}\bigg] \\=-\dfrac {1}{2}[0-1]=\dfrac {1}{2}\)

Illustration Questions

Evaluate  \(I=\displaystyle\int\limits_{1/\pi}^{2/\pi}\dfrac {sin\,( {1}/{x})}{x^2}\,dx\)

A 4

B 1

C –6

D 18

×

\(I=\displaystyle\int\limits_{1/\pi}^{2/\pi}\dfrac {sin\dfrac {1}{x}}{x^2}\,dx\)

Observe that \(\dfrac {1}{x}\) and  \(\dfrac {d}{dx}\dfrac {1}{x}=\dfrac {-1}{x^2}\)  are both there in integrand function.

put \(\dfrac {1}{x}=t\) 

\(\Rightarrow\dfrac {-1}{x^2}\;dx=dt\)

\(\Rightarrow \dfrac {1}{x^2}dx=-dt\)

\(I=-\bigg[\int sin\,t\;dt\bigg]_{x= {1}/{\pi}}^{x= {2}/{\pi}}\)            

 

\(=\bigg[cos\;t \bigg]_{x={1}/{\pi}}^{x={2}/{\pi}}\)  

put back \(t=\dfrac {1}{x}\) and then put the limits.

\(=\bigg[cos\;\dfrac {1}{x} \bigg]_{{1}/{\pi}}^{{2}/{\pi}}\)

\(=cos\dfrac {\pi}{2}-cos\;\pi\)

= 0 – (–1) = 1

 

 

Evaluate  \(I=\displaystyle\int\limits_{1/\pi}^{2/\pi}\dfrac {sin\,( {1}/{x})}{x^2}\,dx\)

A

4

.

B

1

C

–6

D

18

Option B is Correct

Evaluating the Definite Integral by Substitution

Method 2

If 'g' is continuous in [a, b] and 'f' is continuous on the range of u = g(x) then

\(\int\limits_a^b f\;[g(x)]\,g'(x)\,dx=\int\limits_{g(a)}^{g(b)}f(x)dx \)

Proof:

consider,

\(\int\limits_a^b f\;[g(x)]\,g'(x)\,dx\)

\(\text{When},\, x=a\rightarrow \,g(x)=g(a)\\ \text{When},\, x=b\rightarrow \,g(x)=g(b)\\\)   (change two limits according to new variable)

\(\therefore\) \(\int\limits_{g(a)}^{g(b)} f(t)dt= \int\limits_{g(a)}^{g(b)} f(x)dx\)

  • We do not have to go to the original variable x.

Illustration Questions

Evaluate I = \(\int\limits_0^{\pi/2} \sqrt{sin\,x} \,cos^5x\,dx\)  

A \(\dfrac{98}{17}\)

B \(\dfrac{64}{231}\)

C \(\dfrac{5}{9}\)

D \(-\dfrac{3}{4}\)

×

I = \(\int\limits_0^{\pi/2} \sqrt{sin\,x} \,cos^5x\,dx\)

\(\int\limits_0^{\pi/2} \sqrt{sin\,x} ×\,cos^4x × cosx\,dx\)

observe that function \(sin\,x\) and  \(\dfrac{d}{dx}sin x = cos x\) both are present in the integrand.

put \(sin x = t \)

\(\Rightarrow\) \(cos x \,dx = dt\)

change the limits

when x = 0 \(\rightarrow\) sin 0 = t

 \(\Rightarrow\) t = 0

when x = \(\pi\over2\)\(\rightarrow\) sin\(\pi\over2\)

\(\Rightarrow\)t = 1

\(\therefore\) I = \(\int\limits_0^{1} \sqrt{t} (1–t^2)^2\,dt\)                (\(\because cos^2x=1–t^2\))

\(\int\limits_0^{1} \sqrt{t} (1+t^4–2t^2)\,dt\)

\(\int\limits_0^{1}( t^{1/2} + t^{9/2}–2t^{5/2})dt\)

\(=\bigg[\dfrac{t^{3/2}}{3/2}\,+\dfrac{t^{11/2}}{11/2} - \dfrac{2t^{7/2}}{7/2}\bigg]^1_0\)

\(=\dfrac{2}{3}+\dfrac{2}{11} –\dfrac{4}{7}\\=\dfrac{64}{231}\)

Evaluate I = \(\int\limits_0^{\pi/2} \sqrt{sin\,x} \,cos^5x\,dx\)  

A

\(\dfrac{98}{17}\)

.

B

\(\dfrac{64}{231}\)

C

\(\dfrac{5}{9}\)

D

\(-\dfrac{3}{4}\)

Option B is Correct

Integrals of Symmetric Function

  • If \('f'\) is an odd function i.e. \(f(-x) = - f(x)\) then \(\int\limits_{-a}^{a}f(x)\;dx=0\).
  • If the integral is an odd function and limits are negative of each other then its value is 0.
  • Whenever the lower & upper limits are negative of each other, test the integrand function. If it is odd the integral will be 0.

Illustration Questions

The value of \(I=\int\limits_{-\pi/2}^{\pi/2}\;sin\,x\;cos^2\,x\;dx\) has the value.

A \(\dfrac {\pi}{2}\)

B 0

C \(-\dfrac {\pi}{2}\)

D 7

×

\(I=\int\limits_{-\pi/2}^{\pi/2}\;sin\,x\;cos^2\,x\;dx\)

Since the lower limit and upper limit are negative of each other, test the integrand function.

\(f(x)=sin\,x\;cos^2x\)

\(\Rightarrow f(-x)=sin(-x)\;cos^2(-x)\)

\(=-sin\,x\;cos^2x\)

\(=-f(x)\)

\(\Rightarrow f(-x)=-f(x)\\\Rightarrow f \,\,\text{is odd}\)

 

\(\therefore\; I=\int\limits_{-\pi/2}^{\pi/2}\;sin\,x\;cos^2\,x\;dx=0\)

The value of \(I=\int\limits_{-\pi/2}^{\pi/2}\;sin\,x\;cos^2\,x\;dx\) has the value.

A

\(\dfrac {\pi}{2}\)

.

B

0

C

\(-\dfrac {\pi}{2}\)

D

7

Option B is Correct

Integrals of Symmetric Function

  • If \('f'\) is an even function \(\Rightarrow f(-x)=f(x)\), then 

\(I=\int_{-a}^{a}\;f(x)\;dx=2\int_0^af(x)\;dx\) ...(1)

  • f is even therefore graph is symmetric about y-axis

  • If the lower and upper limits are negative of each other, test the integrand function, if it is even, then use equation (1) to simplify the calculation.

Illustration Questions

Evaluate \(I=\int\limits_{-2}^{2}(2x^4+x^2)\;dx\)

A \(\dfrac {47}{6}\)

B \(\dfrac {-5}{3}\)

C \(\dfrac {464}{15}\)

D \(\dfrac {558}{13}\)

×

\(I=\int\limits_{-2}^{2}(2x^4+x^2)\;dx\)

Since lower and upper limits are negative of each other, test the integrand function.

\(f(x)=2x^4+x^2\)

\(\Rightarrow f(-x)=2×(-x)^4+(-x)^2\)

\(\Rightarrow 2×(-1)^4x^4+2×(-1)^2x^2=2x^4+x^2=f(x)\)

\(\therefore\; f(-x)=f(x)\)

\(\rightarrow f\) is even function.

\(I=\int\limits_{-2}^{2}(2x^4+x^2)\;dx=2\int_0^2(2x^4+x^2)\;dx\)

\(=2\left [ \dfrac {2x^5}{5}+\dfrac {x^3}{3} \right]_0^2\)

\(=2\left [ \dfrac {2}{5}×32+\dfrac {8}{3} \right]\)

\(=2\left [ \dfrac {64}{5}+\dfrac {8}{3} \right]\)

\(=2\left [ \dfrac {192+40}{15} \right]\\=2×\dfrac {232}{15}\\=\dfrac {464}{15}\)

Evaluate \(I=\int\limits_{-2}^{2}(2x^4+x^2)\;dx\)

A

\(\dfrac {47}{6}\)

.

B

\(\dfrac {-5}{3}\)

C

\(\dfrac {464}{15}\)

D

\(\dfrac {558}{13}\)

Option C is Correct

The Value of Definite Integral in Terms of another Integral whose Value is Given

  • Sometimes we are given the value of a definite integral and we are asked about another integrand which on suitable substitution can be converted to a multiple of the given integral.
  • e.g,

Suppose \(\int\limits_0^4f(x)\;dx=5\) and we are asked to find the value of  \(I=\int\limits_0^{16}\;\dfrac {f(\sqrt x)}{\sqrt x}\;dx\)

we put \(\sqrt x=t\) 

\(\Rightarrow \dfrac {1}{2\sqrt x}dx=dt\)

\(\Rightarrow \dfrac {1}{\sqrt x}dx=2\;dt\)

\(\therefore\;I=\int\limits_0^4\;f(t)×2dt\)

\(=2\int\limits_0^4\;f(t)\;dt\\=2\int\limits_0^4\;f(x)\;dx\\=2×5=10\)

Illustration Questions

If \('f'\) is continuous and\(\int\limits_0^8\;f(x)\;dx=5\) then find the value of  \(I=\int\limits_0^2x^2\;f(x^3)dx\)

A \(\dfrac {4}{7}\)

B \(\dfrac {5}{3}\)

C –5

D 5

×

\(I=\int\limits_0^2x^2\;f(x^3)dx\)

Observe that \(x^3\) and its derivative \(3x^2\) are both there in the integrand function.

Put \(x^3=t\) 

\(\Rightarrow\;3x^2\;dx=dt\)

\(\Rightarrow x^2\;dx=\dfrac {dt}{3}\)

when \(x=0\rightarrow t=0\)

when \(x=2\rightarrow t=8\)

\(\therefore I=\dfrac {1}{3}\int\limits_0^8\;f(t)\;dt\)

\(=\dfrac {1}{3}\int\limits_0^8\;f(x)\;dx\\=\dfrac {1}{3}×5=\dfrac {5}{3}\)

If \('f'\) is continuous and\(\int\limits_0^8\;f(x)\;dx=5\) then find the value of  \(I=\int\limits_0^2x^2\;f(x^3)dx\)

A

\(\dfrac {4}{7}\)

.

B

\(\dfrac {5}{3}\)

C

–5

D

5

Option B is Correct

Important Property of a Definite Integral

\(\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx\)

  • If the lower limit is 0, there we can replace \(x\)in the integral function by \((a-x)\) everywhere where 'a' is the upper limit. The value of integral remain unchanged.
  • Graphically this property says that 

Shaded area A, can be written in two ways.

  1. \(A_1=\int\limits_0^af(x)\;dx\)
  2. If we take the height of rectangle in reverse order we get the same area as \(A_1=\int_0^af(a-x)dx\)
  • This property often helps in simplifying trigonometric integrals, or expressing one integral in terms of other.

Proof:

\(I=\int\limits_0^a\;f(x)\;dx\)

put  \(a-x=t\)

\(\Rightarrow \,–\,dx=dt\)

when  \(x=0\rightarrow t=a\) 

when \(x=a\rightarrow t=0\)

 \(I=\int\limits_a^0\;f(a-t)×-dt\)

\(I=\int\limits_0^a\;f(a-t)\;dt\\ =\int\limits_0^a\;f(a-x)\;dt\)

\(\therefore \; I=\int\limits_0^a\;f(x)\;dt\\ =\int\limits_0^a\;f(a-x)\;dt\)

Illustration Questions

Which of the following statement is true?

A \(\displaystyle\int\limits_0^{\pi/4}f\;(cos\,x)\;dx = \displaystyle\int\limits_0^{\pi/4}f\;(sin\,x )\;dx\)

B \(\displaystyle\int\limits_0^{\pi/2}f\;(tan\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(cot\,x )\;dx\)

C \(\displaystyle\int\limits_0^{\pi/2}f\;(sin\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(tan\,x )\;dx\)

D \(\displaystyle\int\limits_0^{\pi/4}f\;(tan\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(cot\,x )\;dx\)

×

Apply  \(\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx\)

We know that \(tan\left (\dfrac {\pi}{2}-x\right)=cot\;x\)

\(\therefore\) Only 'B' option is correct.

Which of the following statement is true?

A

\(\displaystyle\int\limits_0^{\pi/4}f\;(cos\,x)\;dx = \displaystyle\int\limits_0^{\pi/4}f\;(sin\,x )\;dx\)

.

B

\(\displaystyle\int\limits_0^{\pi/2}f\;(tan\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(cot\,x )\;dx\)

C

\(\displaystyle\int\limits_0^{\pi/2}f\;(sin\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(tan\,x )\;dx\)

D

\(\displaystyle\int\limits_0^{\pi/4}f\;(tan\,x)\;dx = \displaystyle\int\limits_0^{\pi/2}f\;(cot\,x )\;dx\)

Option B is Correct

Evaluation of Some Definite Integral using the Property  \(\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx\)

Important Property of a Definite Integral

\(\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx\)

  • If the lower limit is 0, there we can replace \(x\)in the integrand function by \((a-x)\) everywhere, where 'a' is the upper limit. The value of integral remain unchanged.
  • Graphically this property says that 

Shaded area A, can be written in two ways:.

  1. \(A_1=\int\limits_0^af(x)\;dx\)
  2. If we take the height of rectangle in reverse order we get the same area as \(A_1=\int_0^af(a-x)dx\)
  • This property often helps in simplifying trigonometric integrals, or expressing one integral in terms of other.

Proof:

\(I=\int\limits_0^a\;f(x)\;dx\rightarrow\)put  \(a-x=t\)

\(\Rightarrow-dx=dt\)

when  \(x=0\rightarrow t=a\) 

when \(x=a\rightarrow t=0\)

 \(I=\int\limits_a^0\;f(a-t)×-dt\)

\(I=\int\limits_0^a\;f(a-t)\;dt =\int\limits_0^a\;f(a-x)\;dt\)

\(\int\limits_0^a\;f(x)\;dt=\int\limits_0^a\;f(a-x)\;dt\)

After applying this property when we add the original integral and that obtained after the property, we get an easy integrand to evaluate.

Illustration Questions

Evaluate \(I=\int\limits_0^1x(1-x)^{10}\;dx\)

A \(\dfrac {4}{5}\)

B \(\dfrac{1}{121}\)

C \(\dfrac{-5}{72}\)

D \(\dfrac {1}{132}\)

×

Apply  \(\int\limits_0^af(x)dx=\int\limits_0^af(a-x)dx\)

\(\Rightarrow\;\int\limits_0^1x(1-x)^{10}\;dx\)

\(\Rightarrow\;\int\limits_0^1(1-x)(1-(1-x))^{10}\;dx\)

\(\Rightarrow\;\int\limits_0^1x^{10}(1-x)\;dx\)

\(\Rightarrow\;\int\limits_0^1(1-x)\;x^{10}\;dx\)

\(\Rightarrow\;\int\limits_0^1x^{10}-x^{11}\;dx\)

\(= \bigg[\dfrac {x^{11}}{11}-\dfrac {x^{12}}{11}\bigg]_0^1\)

\(= \dfrac {1}{11}-\dfrac {1}{12}\\=\dfrac {1}{132}\)

Evaluate \(I=\int\limits_0^1x(1-x)^{10}\;dx\)

A

\(\dfrac {4}{5}\)

.

B

\(\dfrac{1}{121}\)

C

\(\dfrac{-5}{72}\)

D

\(\dfrac {1}{132}\)

Option D is Correct

Some Integrals where Substitution is not So Obvious

Some substitution which are not obvious are

  1. \(\int\limits_0^{\pi/4}\;sin^n\;dx\) where n is odd 

put \(cosx = t\)

     2.    \(\int\limits_0^{\pi/2}\;cos^nx\;dx\) where n is odd

put \(sinx=t\)

    3.   \(\int\limits_0^{\pi/4}\;sec^nx\;dx\) where  n is even 

\(\text { put}\, tan \,x\, = t\)

    4.   \(\int\limits_0^{\pi/4}\;cosec^nx\;dx\) where n is even

put \( cot\, x = t\)

Illustration Questions

Evaluate \(I=\int\limits_0^{\pi/4}\;sec^6x\;dx\)

A \(\dfrac {28}{15}\)

B \(\dfrac {5}{7}\)

C \(\dfrac {-28}{9}\)

D \(\dfrac {11}{4}\)

×

\(I=\int\limits_0^{\pi/4}\;sec^6x\;dx\)

\(I=\underbrace{\int\limits_0^{\pi/4}\;(sec^4x)}_\text{expressed in terms of tan x}×sec^2x\;dx\)

Now, \(\dfrac {d}{dx}(tanx)=sec^2x\)

Put \(tan x = t \)

\(sec^2x\;dx=dt\)

when \(x = 0 \rightarrow t = 0\)

when \(x = \dfrac {\pi}{4} \rightarrow t = 1\)

\(I=\int\limits_0^{1}\;(1+t^2)^2\;dt\)

\(I=\int\limits_0^{1}\;(1+2t^2+t^4)\;dt\)

\(=\Bigg[t+\dfrac {2t^3}{3}+\dfrac {t^5}{5}\Bigg]_0^1\)

\(=1+\dfrac {2}{3}+\dfrac {1}{5}\)

\(=\dfrac {15+10+3}{15}\\=\dfrac {28}{15}\)

Evaluate \(I=\int\limits_0^{\pi/4}\;sec^6x\;dx\)

A

\(\dfrac {28}{15}\)

.

B

\(\dfrac {5}{7}\)

C

\(\dfrac {-28}{9}\)

D

\(\dfrac {11}{4}\)

Option A is Correct

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