Question

$$\underset{0}{\overset{\pi }{\int }}\frac{1}{a+b\cos x}dx=$$

पर्याय

• $$\frac{\pi }{\sqrt{a^2-b^2}}$$

• $$\frac{\pi }{ab}$$

• $$\frac{\pi }{a^2+b^2}$$

• (a + b) π

Answer 1

$$\frac{\pi }{\sqrt{a^2-b^2}}$$

We have

$$I={\int }_0^{\pi }\frac{1}{a+b\cos x}dx$$

$$={\int }_0^{\pi }\frac{1}{a+b\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}}dx$$

$$={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{a(1+{\tan }^2\frac{x}{2})+b(1-{\tan }^2\frac{x}{2})}dx$$

$$={\int }_0^{\pi }\frac{1+{\tan }^2\frac{x}{2}}{(a+b)+(a-b){\tan }^2\frac{x}{2}}dx$$

$$={\int }_0^{\pi }\frac{{\sec }^2\frac{x}{2}}{(a+b)+(a-b){\tan }^2\frac{x}{2}}dx$$

$$\text{Putting}\tan \frac{x}{2}=t$$

$$\Rightarrow \frac{1}{2}{\sec }^2\frac{x}{2}dx=dt$$

$$\Rightarrow {\sec }^2\frac{x}{2}dx=2dt$$

$$Whenx\to 0;t\to 0$$

$$andx\to \pi ;t\to \mathrm{\infty }$$

$$\therefore I={\int }_0^{\mathrm{\infty }}\frac{2dt}{(a+b)+(a-b)t^2}$$

$$=\frac{2}{a-b}{\int }_0^{\mathrm{\infty }}\frac{1}{\left(\frac{a+b}{a-b}\right)+t^2}dt$$

$$=\frac{2}{(a-b)}{\int }_0^{\mathrm{\infty }}\frac{1}{{\left(\sqrt{\frac{a+b}{a-b}}\right)}^2+t^2}dt$$

$$=\frac{2}{(a-b)}\times \sqrt{\frac{a-b}{a+b}}{[{\tan }^{-1}\frac{t}{\sqrt{\frac{a+b}{a-b}}}]}_0^{\mathrm{\infty }}$$

$$=\frac{2}{\sqrt{a^2-b^2}}[\frac{\pi }{2}-0]$$

$$=\frac{2}{\sqrt{a^2-b^2}}\left[\frac{\pi }{2}\right]$$

$$=\frac{\pi }{\sqrt{a^2-b^2}}$$