# more integrals

• October 16th 2011, 10:49 AM
Random Variable
more integrals
Don't hide solutions.

1) $\int_{0}^{\frac{\pi}{2}}} \frac{\arctan(\sin x)}{\sin x} \ dx$

2) $\int^{\infty}_{0} \frac{\sin x \ln x}{x} \ dx$

3) $\int_{0}^{\frac{\pi}{2}}} \ln(\cos x) \ln(\sin x) \cos x \sin x \ dx$
• October 16th 2011, 12:34 PM
Random Variable
Re: more integrals
• October 17th 2011, 03:36 PM
uniquesailor
Re: more integrals
• October 17th 2011, 11:46 PM
FernandoRevilla
Re: more integrals
We have $\lim_{x\to 0^+}\frac{\arctan(\lambda \sin x)}{\sin x}=\ldots=\lambda$ , so the following function is well defined (by means of a convergent integral):

$I:[0,+\infty)\to \mathbb{R},\;\;I(\lambda)=\int_0^{\pi/2}\frac{\arctan (\lambda \sin x)}{\sin x}dx$

Then $I'(\lambda)=\int_0^{\pi/2}\frac{dx}{1+\lambda^2\sin^2 x}$

Using the substitution $t=\tan x$:

$I'(\lambda)=\int_0^{+\infty}\frac{ \frac{dt}{1+t^2} }{ 1+\frac{\lambda^2 t^2}{1+t^2} }=\int_0^{+\infty}\frac{dt}{(1+\lambda^2)t^2+1}$

Using the substitution $u=\sqrt{1+\lambda^2}t$:

$I'(\lambda)=\frac{1}{\sqrt{1+\lambda^2}}\int_{0}^{ +\infty}\frac{du}{u^2+1}=\frac{1}{\sqrt{1+\lambda^ 2}}\dfrac{\pi}{2}$

Integrating both sides:

$I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|+C$

For $\lambda=0$ we have $0=0+C$ that is, $C=0$. Then, $I(\lambda)=\dfrac{\pi}{2}\log |\lambda+\sqrt{\lambda^2+1}|$.

As a consequence:

$\int_0^{\pi/2}\frac{\arctan (\sin x)}{\sin x}dx=I(1)=\dfrac{\pi}{2}\log (1+\sqrt{2})=\boxed{\dfrac{\pi}{2}\sinh^{-1}(1)}$
• December 5th 2011, 12:10 AM
simplependulum
Re: more integrals
Consider $ab = \frac{1}{2}[a^2 + b^2 - (a-b)^2]$ , we have

$\ln[\sin(x)]\ln[\cos(x)] = \frac{1}{2} \left( \ln^2[\sin(x)] + \ln^2[\cos(x)] - \ln^2[\tan(x)] \right)$

$\int_0^{\frac{\pi}{2}}\ln^2[\sin(x)] \sin(x)\cos(x) ~dx$

$= \int_0^{\frac{\pi}{2}}\ln^2[\cos(x)]\sin(x)\cos(x) ~dx = \int_0^1 x\ln^2(x)~dx = \frac{1}{4}$

$\int_0^{\frac{\pi}{2}}\ln^2[\tan(x)] \sin(x)\cos(x) ~dx$

$= \int_0^{\infty} \frac{x \ln^2(x)}{(x^2+1)^2}~dx$

$= 2 \int_0^1 \frac{x \ln^2(x)}{(x^2+1)^2}~dx$

$= \left[ \frac{x^2\ln^2(x)}{x^2+ 1}\right]_0^1 - 2\int_0^1 \frac{x\ln(x)}{x^2+1}~dx = \frac{\pi^2}{24}$

so the answer is $\frac{1}{4} - \frac{\pi^2}{48}$
• March 4th 2012, 11:19 PM
sbhatnagar
Re: more integrals
Quote:

Originally Posted by uniquesailor

Can you explain what is 'MAZ identity'?