Help to Evaluate an Integral

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.

#Calculus #Integration #HelpMe! #MathProblem

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

First, 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 - 7 years, 6 months ago

its an awesome solution ..!!

gopal chpidhary - 7 years, 6 months ago

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 - 7 years, 6 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.