Find Integral of sin(x) / cos(x)^2 - Homework Help

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Homework Statement



I would just like some help on how to find the integral of:
sin(x) / cos(x)^2



Homework Equations



integration by parts?
uv-integral of v(du)


The Attempt at a Solution


I tried using integration of parts with u = cos(x)^2 and dv = sin(x) but I just got myself in a bigger mess.

I noticed that sin(x) / cos(x)^2 was equal to tan(x)sec(x). Not sure if that helps though...

Help please?!
 
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I'd go for a substitution.
 
Try the substitution cos(x)^2.
 
what's integral of [(-2*u)/(1-u^2)]/-2

or

try to see 1/x connection
 
Last edited:
\int \frac{sinx}{cos^2x} dx

\equiv \int \frac{sinx}{cosx} \times \frac{1}{cosx}dx

What is another way to write \frac{sinx}{cosx}?
and similarly \frac{1}{cosx}?


Should be pretty standard after you see it.
 
Got it. Thank you everyone.
 
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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