Improving Trigonometric Integration

ada0713
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Homework Statement



\int\frac{secx}{(tanx)^2}dx


The Attempt at a Solution


I tried all the u subs u=tanx and u=secx
but neither worked.
Should I used other methods?
Please help me with the start!
 
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ada0713 said:

Homework Statement



\int\frac{secx}{(tanx)^2}dx


The Attempt at a Solution


I tried all the u subs u=tanx and u=secx
but neither worked.
Should I used other methods?
Please help me with the start!

You might try reducing this to sine and cosine. sec x= 1/cos x and tan x = sin x/cos x so this is
\int \frac{1}{cos x}\frac{cos^2 x}{sin^2 x} dx= \int \frac{cos x}{sin^2 x} dx


Anything come to mind now?
 
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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