IIT-JEE 1985Question 1
Evaluate: $\displaystyle\int\sqrt{\frac{1-\sqrt x}{1+\sqrt x}}\,dx$.
Indefinite Integration
Answer$-2\sqrt{1-x}+\cos^{-1}\sqrt x+\sqrt x\sqrt{1-x}+C$
Solution
Concept. With $\sqrt x=\cos\theta$ the root becomes $\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\tan\frac\theta2$, and everything reduces to powers of $\cos\theta$. Formulas used. (1) $x=\cos^2\theta$ ($0\le\theta\le\frac\pi2$): $dx=-2\cos\theta\sin\theta\,d\theta$(2) $\tan\frac\theta2=\frac{1-\cos\theta}{\sin\theta}$(3) $2\cos^2\theta=1+\cos2\theta$Step 1 - Substitute. By (1), (2), $$\int\frac{1-\cos\theta}{\sin\theta}\cdot(-2\cos\theta\sin\theta)\,d\theta=\int\big(-2\cos\theta+2\cos^2\theta\big)\,d\theta.$$ Step 2 - Integrate. By (3), $$\int\big(-2\cos\theta+1+\cos2\theta\big)d\theta=-2\sin\theta+\theta+\sin\theta\cos\theta+C.$$ Step 3 - Back to $x$. $\cos\theta=\sqrt x$, $\sin\theta=\sqrt{1-x}$, $\theta=\cos^{-1}\sqrt x$: $$-2\sqrt{1-x}+\cos^{-1}\sqrt x+\sqrt x\sqrt{1-x}+C.$$ Answer: $-2\sqrt{1-x}+\cos^{-1}\sqrt x+\sqrt x\sqrt{1-x}+C$