# Integration by Substitution Method

## Integration by Substitution Method

When the variable in a defined integral is changed, the substitution in terms of new variable should be effected at three places.

i. In the integration

ii. In the differential, say dx

iii. In the limits

Evaluate: $$\int\limits_{0}^{\pi /2}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right).dx}$$.

Solution: $$\int\limits_{0}^{\pi /2}{\left( \sqrt{\tan x}+\sqrt{\cot x} \right).dx}$$,

= $$\int\limits_{0}^{\pi /2}{\left( \sqrt{\frac{\sin x}{\cos x}}+\sqrt{\frac{\cos x}{\sin x}} \right).dx}$$,

= $$\int\limits_{0}^{\pi /2}{\left( \frac{\sin x+\cos x}{\sqrt{\sin x\cos x}} \right).dx}$$,

= $$\sqrt{2}\int\limits_{0}^{\pi /2}{\frac{1}{\sqrt{2\sin x\cos x}}}.d(-\cos x+\sin x).dx$$,

= $$\sqrt{2}\int\limits_{0}^{\pi /2}{\frac{1}{\sqrt{1-{{(-\cos x+\sin x)}^{2}}}}}.d(-\cos x+\sin x).dx$$,

Let -cosx + sinx = t then

x = 0 → t = -1 and x = π/2 → t = 1

$$=\sqrt{2}\int\limits_{-1}^{1}{\frac{1}{\sqrt{1-{{t}^{2}}}}}.dt$$,

$$=\sqrt{2}\left[ {{\sin }^{-1}}t \right]_{-1}^{1}$$,

$$=\sqrt{2}[{{\sin }^{-1}}1-{{\sin }^{-1}}(-1)]$$,

$$=\sqrt{2}\left( \frac{\pi }{2}-(-\frac{\pi }{2}) \right)$$,

$$=\sqrt{2}\pi$$.

Evaluate: $$\int\limits_{0}^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$$.

Solution: $$\int\limits_{0}^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$$,

Put $${{\sin }^{-1}}x=t\Rightarrow \frac{1}{\sqrt{1-{{x}^{2}}}}dx=dt$$,

x = 0 → t = 0 and x = ½ → t = π/6

$$\int\limits_{0}^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx=\int\limits_{0}^{\pi /2}{t.\sin t.dt}$$,

= $$\left( t\int{\sin t.dt} \right)_{0}^{-\pi /6}-\int\limits_{0}^{\pi /6}{1(-\cos t)dt}$$,

= $$(-t\cos t)_{0}^{\pi /6}+(\sin t)_{0}^{\pi /6}$$,

= $$\frac{\pi }{6}\left( \frac{-\sqrt{3}}{2} \right)-0+\frac{1}{2}-0$$,

= $$\frac{1}{2}-\pi \frac{\sqrt{3}}{12}$$.