Help to Evaluate an Integral

I Need Help in evaluating this integral

0π2θcosθdθsinθ+sin3θ\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.

Note by Ghanashyam Chakravarthi
5 years, 11 months ago

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8 votes

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First, use IBP and we'll end up with 0π2xcosxsinx+sin3xdx=π4log20π2(logsinxlog(1+sin2x)2dx) \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 0π2logsinxdx=π2log2\int_{0}^{\frac{\pi}{2}} \log{\sin{x}}\, dx = -\frac{\pi}{2}\,\log{2}

we have 0π2xcosxsinx+sin3xdx=π4log2+0π2log(1+sin2x)2dx(0)\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)=0π2log(1+sin2x+ksin2x)2dx(1)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:

Ik=0π2sin2xdx1+sin2x+ksin2x\frac{\partial I}{\partial k} = \int_{0}^{\frac{\pi}{2}} \frac{\sin^{2}{x}\, dx}{1+\sin^{2}{x}+k\, \sin^{2}{x}}     I(k)=0π21k+1(1sec2x1+(k+2)tan2x)dx\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 =xk+1tan1(k+2tanx)(k+1)k+20π2=π2k+2(11k+2)=\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)=π2(log(k+1)log(k+21k+2+1))+C=π2log(2(k+2+1)2)+CI(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 I(k)=π2log(2(k+2+1)2)+C(2)\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)=0I(-1) = 0 and from (2),
I(1)=3π2log2+CI(-1) = \frac{3\,\pi}{2}\log{2}+C     C=3π2log2\implies C= - \frac{3\,\pi}{2}\log{2} The required integral is I(0)=0π2log(1+sin2x)dx=π2log(2(2+1)2)3π2log2I(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 0π2xcosxdxsinx+sin3x=π4log2+π4log(2(2+1)2)3π4log2\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} =π4log(3+222)= \boxed{{\frac{\pi}{4}\log{\left(\frac{3+2\,\sqrt{2}}{2}\right)}}}

gopinath no - 5 years, 11 months ago

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its an awesome solution ..!!

gopal chpidhary - 5 years, 11 months ago

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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)?

John Ashley Capellan - 5 years, 11 months ago

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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.

Ghanashyam Chakravarthi - 5 years, 11 months ago

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

Biswaroop Roy - 5 years, 11 months ago

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

Ghanashyam Chakravarthi - 5 years, 11 months ago

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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?

Biswaroop Roy - 5 years, 11 months ago

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@Biswaroop Roy How did you get (pi/2) ln(tan(pi/2)/tan 0)) ? From there on its correct.

Ghanashyam Chakravarthi - 5 years, 11 months ago

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@Ghanashyam Chakravarthi That is very simple. from the denominator take sine theta common. put cos^2(theta)
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 - 5 years, 11 months ago

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@Biswaroop Roy but 1- sin^2 (theta) is cos^2(theta) !!!

Ghanashyam Chakravarthi - 5 years, 11 months ago

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you can solve this by using Wallis fomula

Mark Relosa - 5 years, 11 months ago

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hey can you tell me how..

Ghanashyam Chakravarthi - 5 years, 11 months ago

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