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

Last updated date: 25th Mar 2023

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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 \[{\tan ^{ - 1}}(\cot \alpha )\].

__Complete step-by-step answer:__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 \[{\tan ^{ - 1}}(\cot \alpha )\].

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