question_answer

Evaluate \[\int_{0}^{\pi }{\frac{dx}{3+2\sin x+\cos x}.}\]

Answer:

Let \[l=\int_{0}^{\pi }{\frac{dx}{3+2\sin x+\cos x}}\] \[=\int_{0}^{\pi }{\frac{dx}{3+2\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}+\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}\] \[\left[ \because \sin x=\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}},\cos x=\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}} \right]\] \[=\int_{0}^{\pi }{\frac{dx}{\frac{3+3{{\tan }^{2}}\frac{x}{2}+4\tan \frac{x}{2}+1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}\] \[=\int_{0}^{\pi }{\frac{\left( 1+{{\tan }^{2}}\frac{x}{2} \right)\,dx}{2{{\tan }^{2}}\frac{x}{2}+4\tan \frac{x}{2}+4}}=\int_{0}^{\pi }{\frac{{{\sec }^{2}}\frac{x}{2}}{2{{\tan }^{2}}\frac{x}{2}+4\tan \frac{x}{2}+4}}\]Put \[\tan \frac{x}{2}=t\Rightarrow {{\sec }^{2}}\frac{x}{2}dx=2dt\] Also when \[x=0,\]then \[t=0\]and, when \[x=\pi ,\]then \[t=\infty \] \[\therefore \]      \[l=\int_{0}^{\infty }{\frac{2dt}{2{{t}^{2}}+4t+4}}=\int_{0}^{\infty }{\frac{dt}{{{t}^{2}}+2t+2}}\] \[=\int_{0}^{\infty }{\frac{dt}{{{(t+1)}^{2}}+1}}=[{{\tan }^{-\,1}}(t+1)_{0}^{\infty }]\] \[={{\tan }^{-\,1}}\infty -{{\tan }^{-\,1}}1=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}\]