I Need Help in evaluating this integral

\(\int\limits_0^\frac{\pi}{2} \frac {\theta cos \theta d\theta } { sin \theta + \sin ^ 3 \theta }\)

None of my teachers were able to solve this, nor my friends. It has a finite answer, as one of my friends checked on a scientific calculator.

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## Comments

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TopNewestFirst, use IBP and we'll end up with \[ \int_{0}^{\frac{\pi}{2}}\dfrac{x \cos{x}}{\sin{x}+\sin^{3}{x}}\, dx = -\frac{\pi}{4}\log{2} - \int_{0}^{\frac{\pi}{2}} \left( \log{\sin{x}} - \frac{\log{(1+\sin^{2}{x})}}{2}\, dx \right) \]

and since \[\int_{0}^{\frac{\pi}{2}} \log{\sin{x}}\, dx = -\frac{\pi}{2}\,\log{2} \]

we have \[\int_{0}^{\frac{\pi}{2}}\dfrac{x \cos{x}}{\sin{x}+\sin^{3}{x}} \, dx = \frac{\pi}{4}\,\log{2} + \int_{0}^{\frac{\pi}{2}} \frac{\log{(1+\sin^{2}{x})}}{2}\, dx \quad\cdots\cdots\cdots\cdots (0)\]

Let \[I(k) = \int_{0}^{\frac{\pi}{2}} \frac{\log{(1+\sin^{2}{x}+k\, \sin^{2}{x})}}{2}\, dx \quad\cdots\cdots\cdots\cdots (1)\]

To evaluate I(k), we can make use of differentiation under the integral sign:

\[\frac{\partial I}{\partial k} = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}{x}\, dx}{1+\sin^{2}{x}+k\, \sin^{2}{x}} \] \[\implies I'(k) = \int_{0}^{\frac{\pi}{2}} \frac{1}{k+1}\left(1-\frac{\sec^{2}{x}}{1+(k+2)\tan^{2}x}\right)\, \, dx \] \[=\frac{x}{k+1}-\frac{\tan^{-1}{(\sqrt{k+2}\,\tan{x})}}{(k+1)\, \sqrt{k+2}}\bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{2\,k+2}\left(1-\frac{1}{\sqrt{k+2}}\right)\]

Integrating w.r.t. k, we get:

\[I(k)= \frac{\pi}{2}\left(\log{(k+1)}-\log{\left(\frac{\sqrt{k+2}-1}{\sqrt{k+2}+1}\right)}\right)+C=\frac{\pi}{2}\, \log{\left(2\, (\sqrt{k+2}+1)^{2}\right)}+C\] \[\therefore I(k) = \frac{\pi}{2}\, \log{\left(2\, (\sqrt{k+2}+1)^{2}\right)}+C \quad\cdots\cdots\cdots\cdots (2)\]

To find C, substitute k=-1. From (1), \[I(-1) = 0\] and from (2),

\[I(-1) = \frac{3\,\pi}{2}\log{2}+C\] \[\implies C= - \frac{3\,\pi}{2}\log{2}\] The required integral is \[I(0) = \int_{0}^{\frac{\pi}{2}} \log{(1+\sin^{2}{x})}\, dx = \frac{\pi}{2}\, \log{\left(2\, (\sqrt{2}+1)^{2}\right)} - \frac{3\,\pi}{2}\log{2}\]

Substituting in (0), we see that \[\int_{0}^{\frac{\pi}{2}}\dfrac{x \cos{x}\, dx}{\sin{x}+\sin^{3}{x}} = \frac{\pi}{4}\,\log{2} + \frac{\pi}{4}\, \log{\left(2\, (\sqrt{2}+1)^{2}\right)}- \frac{3\,\pi}{4}\log{2}\] \[= \boxed{{\frac{\pi}{4}\log{\left(\frac{3+2\,\sqrt{2}}{2}\right)}}}\]

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its an awesome solution ..!!

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you can solve this by using Wallis fomula

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hey can you tell me how..

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I tried using limits. What's the answer? Is it by any chance 0??

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I don't know the answer, but I don't think its zero.

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Look,at first the sum looks easy,doesn't it? my answer was (pi/2) ln(tan(pi/2)/tan 0)) Now as this tan((pi/2)/tan(0)) is undefined. so lets find limit x->0 tan((pi/2)+x)/tan(x). As this is (infinity)/(infinity) form, we can differentiate on numerator and denominator according to L'Hospital Rule. Doing so, and putting x=0,once the expression is defined, we get the limiting value to be 1. So putting this limiting value in the initial expression, answer is (pi/2)(ln 1) or, (pi/2). 0 or 0. In the end answer is coming out to be 0. Did I go wrong anywhere?

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in place of 1+ sin^2(theta). Then do the required cancellation to get expression as ((theta)/(sin (theta).cos(theta))) inside the integral. Now apply,the rule for definite integral: Integral f(x)from 0 to a=integral f(a-x)from 0 to a. Put the limits from 0 to pi/2.Then you will get the final expression : pi/2(sec^2(theta)d(theta)/tan(theta)) Now you should be able to do the rest. Its simple enough .

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Hello... Where is d(theta)? This is normally written when integrating functions to say reverse the process and of course, finding the antiderivative. Also is theta really before cos(theta)?

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Thanks John. I had forgotten d(theta) in the process of tying in Latex. And yes there is a (theta) before cos(theta). I've made the change.

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