What is the Easiest Way to Solve an Integration Problem Involving Tan(x)?

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Solved. Thanks.
 
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The simplest way to solve that indefinite integral is to realize that

\frac{d}{dx}tan x = sec^{2}x , and try a u-substitution from there...
 
meiso said:
The simplest way to solve that indefinite integral is to realize that

\frac{d}{dx}tan x = sec^{2}x , and try a u-substitution from there...

Yes, sorry, i am stupid. I realized that 2 minutes after posting here. I'm pretty sure i have it now. It's just -(2+u)^{-1} for u=tanx, right? And thanks for the reply.
 
Noo said:
It's just -(2+u)^{-1} for u=tanx, right? And thanks for the reply.

No problem. And, actually, you can set u = tan(x) + 2 to make things even easier. 2 is just a constant, so the derivative of tan(x) is the same as the derivative of tan(x) + 2.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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