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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  $$t=x^2=\dfrac{\pi}{2}$$

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

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

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

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

#### 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.

#### 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.

#### 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.

#### 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 $$\sqrt x=t$$

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

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

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

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

#### 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 $$x=a\rightarrow t=0$$

$$I=\int\limits_a^0\;f(a-t)×-dt$$

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

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

#### 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.

#### 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 $$cot\, x = t$$

#### 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