How to Evaluate Trig Integrals and Simplify with Trig Identities?

[dx/dy=?]

Can someone please help me in evaluating the following integral?

Im really bad with trig functions, and have been trying to work this out for ages.

\int \frac{sin x}{cos^2 x} dx

Is there a way I can evaluate for \cos^2 x ?
Or is there something I am missing altogether?



Also,

Can someone please tell me how:
\int \frac{sec\theta}{cos\theta} dx =\ tan\theta + C
??

 

[TD]

\int \frac{sin x}{cos^2 x} dx

Substitution: y = \cos x \Leftrightarrow dy = - \sin xdx gives:

\int {\frac{{\sin x}}<br /> {{\cos ^2 x}}dx} = \int {\frac{{ - 1}}<br /> {{y^2 }}dy = } \frac{1}<br /> {y} + C = \frac{1}<br /> {{\cos x}} + C

Can someone please tell me how:
\int \frac{sec\theta}{cos\theta} dx =\ tan\theta = C

If you know that :
\sec \theta = \frac{1}{{\cos \theta }}

and

\frac{{d\left( {\tan \theta } \right)}}<br /> {{d\theta }} = \frac{1}<br /> {{\cos ^2 \theta }}

then...

 

[dx/dy=?]

Thanks,

For the first integral \int \frac{sin x}{cos^2 x} dx was thinking more along the lines of multiplying out before evaluatng, to try to obtain a more simple integral to evaluate,
but your way is obviously better.

Is there a way this can be evaluated without substitution?
Or is there no function which has \cos^2 x as its derivative?

Thanks again for the help.

 

[GCT]

why would you want cos^2@ as a derivative?

 

[TD]

Is there a way this can be evaluated without substitution?
Or is there no function which has \cos^2 x as its derivative?

Thanks again for the help.

The easiest way is certainly using that substitution.

I don't really see why you'd want to know an anti-derivative for \cos^2 x, if you do: just compute it, use:
\cos \left( {2x} \right) = 2\cos ^2 x - 1 \Leftrightarrow \cos ^2 x = \frac{{1 + \cos \left( {2x} \right)}}<br /> {2}

I don't see how that would help though, you don't have \cos^2 x in the integral but \frac{1}<br /> {{\cos ^2 x}}, whose anti-derivative is of course \tan x

 

[whozum]

\int \frac{sin x}{cos^2 x} dx = \int \sec x \tan x dx = \sec x + C

 

[dx/dy=?]

Thanks very much for all your help everyone.

Im slowly getting the hang of Calculus, but I am not at all good with derivatives of trig functions.
Ill get there eventually.