IIT-JEE 1986Question 1
Evaluate: $\displaystyle\int_0^{\pi}\frac{x\,dx}{1+\cos\alpha\,\sin x},\ 0<\alpha<\pi$.
Definite Integration
Answer$\dfrac{\pi\alpha}{\sin\alpha}$
Solution
Concept. King's rule removes the factor $x$ ($\sin(\pi-x)=\sin x$). Then $t=\tan\frac x2$ turns the integral into $\int_0^\infty\frac{2\,dt}{(t+\cos\alpha)^2+\sin^2\alpha}$. Formulas used. (1) $\int_0^\pi x\,g(\sin x)\,dx=\frac\pi2\int_0^\pi g(\sin x)\,dx$(2) $t=\tan\frac x2$: $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2\,dt}{1+t^2}$; $x:0\to\pi$ gives $t:0\to\infty$(3) $\int\frac{dt}{(t+c)^2+s^2}=\frac1s\tan^{-1}\frac{t+c}{s}$; and $\tan^{-1}(\cot\alpha)=\frac\pi2-\alpha$ for $0<\alpha<\pi$Step 1 - Use (1). $I=\frac\pi2\int_0^\pi\frac{dx}{1+\cos\alpha\sin x}$. Step 2 - Use (2). $$\int_0^\pi\frac{dx}{1+\cos\alpha\sin x}=\int_0^\infty\frac{2\,dt}{t^2+2t\cos\alpha+1}=\int_0^\infty\frac{2\,dt}{(t+\cos\alpha)^2+\sin^2\alpha}.$$ Step 3 - Use (3). ($\sin\alpha>0$.) $$\frac{2}{\sin\alpha}\Big[\frac\pi2-\tan^{-1}(\cot\alpha)\Big]=\frac{2}{\sin\alpha}\Big[\frac\pi2-\Big(\frac\pi2-\alpha\Big)\Big]=\frac{2\alpha}{\sin\alpha}.$$ Step 4 - Value. $I=\frac\pi2\cdot\frac{2\alpha}{\sin\alpha}=\frac{\pi\alpha}{\sin\alpha}$. Answer: $\dfrac{\pi\alpha}{\sin\alpha}$