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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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 \[\int_{0}^{\frac{\pi}{2}}\dfrac{x \cos{x}\, dx}{\sin{x}+\sin^{3}{x}} = \frac{\pi}{4}\,\log{2} + \frac{\pi}{4}\, \log{\left(2\, (\sqrt{2}+1)^{2}\right)}- \frac{3\,\pi}{4}\log{2}\] \[= \boxed{{\frac{\pi}{4}\log{\left(\frac{3+2\,\sqrt{2}}{2}\right)}}}\] – Gopinath No · 3 years, 8 months ago

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– Gopal Chpidhary · 3 years, 8 months ago

its an awesome solution ..!!Log in to reply

you can solve this by using Wallis fomula – Mark Relosa · 3 years, 8 months ago

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– Ghanashyam Chakravarthi · 3 years, 8 months ago

hey can you tell me how..Log in to reply

I tried using limits. What's the answer? Is it by any chance 0?? – Biswaroop Roy · 3 years, 8 months ago

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– Ghanashyam Chakravarthi · 3 years, 8 months ago

I don't know the answer, but I don't think its zero.Log in to reply

– Biswaroop Roy · 3 years, 8 months ago

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?Log in to reply

– Ghanashyam Chakravarthi · 3 years, 8 months ago

How did you get (pi/2) ln(tan(pi/2)/tan 0)) ? From there on its correct.Log in to reply

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 . – Biswaroop Roy · 3 years, 8 months ago

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– Ghanashyam Chakravarthi · 3 years, 8 months ago

but 1- sin^2 (theta) is cos^2(theta) !!!Log in to reply

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)? – John Ashley Capellan · 3 years, 8 months ago

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– Ghanashyam Chakravarthi · 3 years, 8 months ago

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.Log in to reply