IIT-JEE 1989Question 1
Evaluate $\displaystyle\int\left(\sqrt{\tan x}+\sqrt{\cot x}\right)dx$.
Indefinite Integration
Answer$\sqrt2\tan^{-1}\left(\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt2}\right)+C$
Solution
Concept. Put $t=\sqrt{\tan x}$: the integrand becomes $\frac{2(t^2+1)}{t^4+1}$. Dividing by $t^2$, the numerator is $d\big(t-\frac1t\big)$. Formulas used. (1) $\tan x=t^2$: $\sec^2x\,dx=2t\,dt$, so $dx=\frac{2t\,dt}{1+t^4}$(2) $w=t-\frac1t$: $dw=\big(1+\frac1{t^2}\big)dt$, $t^2+\frac1{t^2}=w^2+2$(3) $\int\frac{dw}{w^2+2}=\frac1{\sqrt2}\tan^{-1}\frac w{\sqrt2}$Step 1 - Substitute. $\sqrt{\cot x}=\frac1t$. By (1), $$\int\Big(t+\frac1t\Big)\frac{2t\,dt}{1+t^4}=2\int\frac{t^2+1}{t^4+1}\,dt.$$ Step 2 - Divide by $t^2$. $$2\int\frac{1+\frac1{t^2}}{t^2+\frac1{t^2}}\,dt=2\int\frac{dw}{w^2+2}\quad\text{by (2)}.$$ Step 3 - Use (3). $\sqrt2\tan^{-1}\frac{w}{\sqrt2}+C$, and $w=t-\frac1t=\sqrt{\tan x}-\sqrt{\cot x}$. Answer: $\sqrt2\tan^{-1}\left(\dfrac{\sqrt{\tan x}-\sqrt{\cot x}}{\sqrt2}\right)+C$