Question

# Evaluate the following integral: $\int_0^\pi {\dfrac{{xdx}}{{1 + \cos \alpha \sin x}}}$ where $0 < \alpha < \pi$.

Hint: Use the properties of definite integrals, that is, $\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} }$ and simplify the expression to obtain the desired result.

Let us assign the integral to a variable I.
$I = \int_0^\pi {\dfrac{{xdx}}{{1 + \cos \alpha \sin x}}} ..........(1)$
We know the formula for definite integral as follows:
$\int_0^a {f(x)dx = \int_0^a {f(a - x)dx} } .........(2)$
Here, $f(x) = \dfrac{x}{{1 + \cos \alpha \sin x}}$ and a = $\pi$.
Hence, using formula in equation (2) to simplify equation (1), we get:
$I = \int_0^\pi {\dfrac{{(\pi - x)dx}}{{1 + \cos \alpha \sin (\pi - x)}}}$
We know that $\sin (\pi - x) = \sin x$, sine is positive in the second quadrant. Hence, we get:
$I = \int_0^\pi {\dfrac{{(\pi - x)dx}}{{1 + \cos \alpha \sin x}}}$
Expanding the numerator, we have:
$I = \pi \int_0^\pi {\dfrac{{dx}}{{1 + \cos \alpha \sin x}}} - \int_0^\pi {\dfrac{{xdx}}{{1 + \cos \alpha \sin x}}}$
The second term in the expression is nothing but I itself, hence, using equation(1), we have:
$I = \pi \int_0^\pi {\dfrac{{dx}}{{1 + \cos \alpha \sin x}}} - I$
Solving for I, we get:
$2I = \pi \int_0^\pi {\dfrac{{dx}}{{1 + \cos \alpha \sin x}}}$
$I = \dfrac{\pi }{2}\int_0^\pi {\dfrac{{dx}}{{1 + \cos \alpha \sin x}}} .........(3)$
Let us consider the integral term alone in equation (3).
$I' = \int_0^\pi {\dfrac{{dx}}{{1 + \cos \alpha \sin x}}} .........(4)$
Now, we know that, $\sin 2x = \dfrac{{2\tan x}}{{1 + {{\tan }^2}x}}$, using this formula for sin(x) in equation (4), we get:
$I' = \int_0^\pi {\dfrac{{\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)dx}}{{1 + {{\tan }^2}\dfrac{x}{2} + \cos \alpha .2\tan \dfrac{x}{2}}}} ........(5)$
Let us use substitution of variables as follows:
$\tan \dfrac{x}{2} = \theta$
${\sec ^2}\dfrac{x}{2}.\dfrac{1}{2}.dx = d\theta$
$\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right)dx = 2d\theta$
The limits also change as follows:
$x \to 0 \Rightarrow \theta \to 0$
$x \to \pi \Rightarrow \theta \to \infty$
Using all this changes in equation (5), we get:
$I' = \int_0^\infty {\dfrac{{2d\theta }}{{1 + {\theta ^2} + 2\theta \cos \alpha }}}$
Expressing the denominator as sum of squares using completing square method, we get:
$I' = \int_0^\infty {\dfrac{{2d\theta }}{{{{(\theta + \cos \alpha )}^2} + (1 - {{\cos }^2}\alpha )}}}$
We know that $1 - {\cos ^2}x = {\sin ^2}x$, hence we have:
$I' = 2\int_0^\infty {\dfrac{{d\theta }}{{{{(\theta + \cos \alpha )}^2} + {{\sin }^2}\alpha }}}$
We know that $\int {\dfrac{{dx}}{{{{(x)}^2} + {a^2}}} = \dfrac{1}{a}\tan {}^{ - 1}\dfrac{x}{a}}$, hence we have:
$I' = \left. {\dfrac{2}{{\sin \alpha }}{{\tan }^{ - 1}}\left( {\dfrac{{\theta + \cos \alpha }}{{\sin \alpha }}} \right)} \right|_0^\infty$
Evaluating the limits, we have:
$I' = \dfrac{2}{{\sin \alpha }}\left( {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {\dfrac{{\cos \alpha }}{{\sin \alpha }}} \right)} \right)$
We know that, $\dfrac{{\cos \alpha }}{{\sin \alpha }} = \cot \alpha$, hence we get:
$I' = \dfrac{2}{{\sin \alpha }}\left( {{{\tan }^{ - 1}}\left( \infty \right) - {{\tan }^{ - 1}}\left( {\cot \alpha } \right)} \right)$
We also know that, ${\tan ^{ - 1}}\left( \infty \right) = \dfrac{\pi }{2}$ and $\cot \alpha = \tan \left( {\dfrac{\pi }{2} - \alpha } \right)$, hence we have:
$I' = \dfrac{2}{{\sin \alpha }}\left( {\dfrac{\pi }{2} - {{\tan }^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{2} - \alpha } \right)} \right)} \right)$
Since, ${\tan ^{ - 1}}\left( {\tan \left( x \right)} \right) = x$, we have:
$I' = \dfrac{2}{{\sin \alpha }}\left( {\dfrac{\pi }{2} - \left( {\dfrac{\pi }{2} - \alpha } \right)} \right)$
Simplifying, we get:
$I' = \dfrac{{2\alpha }}{{\sin \alpha }}.........(6)$
Using equation (6) in equation (3), we get:
$I = \dfrac{\pi }{2} \times \dfrac{{2\alpha }}{{\sin \alpha }}$
$I = \dfrac{{\pi \alpha }}{{\sin \alpha }}$
Hence, the value of the integral is $I = \dfrac{{\pi \alpha }}{{\sin \alpha }}$.

Note: You might make mistake in the integration of $\int {\dfrac{{dx}}{{{{(x)}^2} + {a^2}}} = \dfrac{1}{a}\tan {}^{ - 1}\dfrac{x}{a}}$ formula by missing out the $\dfrac{1}{a}$ term. It is necessary to simplify the integral completely and not leave the answer in terms of ${\tan ^{ - 1}}(\cot \alpha )$.