Question

\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

विकल्प

• \[\frac{\pi}{\sqrt{a^2 - b^2}}\]

• \[\frac{\pi}{ab}\]

• \[\frac{\pi}{a^2 + b^2}\]

• (a + b) π

Answer 1

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

We have

\[I = \int_0^\pi \frac{1}{a + b\cos x} d x\]

\[ = \int_0^\pi \frac{1}{a + b\frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}} d x\]

\[= \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{a\left( 1 + \tan^2 \frac{x}{2} \right) + b\left( 1 - \tan^2 \frac{x}{2} \right)} d x\]

\[ = \int_0^\pi \frac{1 + \tan^2 \frac{x}{2}}{\left( a + b \right) + \left( a - b \right) \tan^2 \frac{x}{2}}dx\]

\[ = \int_0^\pi \frac{\sec^2 \frac{x}{2}}{\left( a + b \right) + \left( a - b \right) \tan^2 \frac{x}{2}}dx\]

\[\text{Putting} \tan\frac{x}{2} = t\]

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

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

\[When\ x \to 0; t \to 0\]

\[and\ x \to \pi; t \to \infty \]

\[ \therefore I = \int_0^\infty \frac{2dt}{\left( a + b \right) + \left( a - b \right) t^2}\]

\[ = \frac{2}{a - b} \int_0^\infty \frac{1}{\left( \frac{a + b}{a - b} \right) + t^2}dt\]

\[= \frac{2}{\left( a - b \right)} \int_0^\infty \frac{1}{\left( \sqrt{\frac{a + b}{a - b}} \right)^2 + t^2}dt\]

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

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

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

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