Evaluate the following integral: \[\int_0^\pi {\dfrac{{xdx}}{{1 + \cos \alpha \sin x}}} \] where \[0 < \alpha < \pi \].
Answer
363k+ views
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.
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 \[{\tan ^{ - 1}}(\cot \alpha )\].
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 \[{\tan ^{ - 1}}(\cot \alpha )\].
Last updated date: 29th Sep 2023
•
Total views: 363k
•
Views today: 8.63k
Recently Updated Pages
What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE
