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
\(\int\limits_{0}^{1/2}{\frac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx=\int\limits_{0}^{\pi /2}{t.\sin t.dt}\),
= \(\left( t\int{\sin t.dt} \right)_{0}^{-\pi /6}-\int\limits_{0}^{\pi /6}{1(-\cos t)dt}\),
= \((-t\cos t)_{0}^{\pi /6}+(\sin t)_{0}^{\pi /6}\),
= \(\frac{\pi }{6}\left( \frac{-\sqrt{3}}{2} \right)-0+\frac{1}{2}-0\),
= \(\frac{1}{2}-\pi \frac{\sqrt{3}}{12}\).