# Evaluating Trig' Integrals

1. Aug 7, 2005

### dx/dy=?

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 Im missing altogether?

Also,

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

Last edited: Aug 7, 2005
2. Aug 7, 2005

### TD

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

$$\int {\frac{{\sin x}} {{\cos ^2 x}}dx} = \int {\frac{{ - 1}} {{y^2 }}dy = } \frac{1} {y} + C = \frac{1} {{\cos x}} + C$$

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

and

$$\frac{{d\left( {\tan \theta } \right)}} {{d\theta }} = \frac{1} {{\cos ^2 \theta }}$$

then...

3. Aug 8, 2005

### 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.

Last edited: Aug 8, 2005
4. Aug 8, 2005

### GCT

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

5. Aug 8, 2005

### TD

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:
$$\cos \left( {2x} \right) = 2\cos ^2 x - 1 \Leftrightarrow \cos ^2 x = \frac{{1 + \cos \left( {2x} \right)}} {2}$$

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

6. Aug 8, 2005

### whozum

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

7. Aug 9, 2005

### dx/dy=?

Thanks very much for all your help everyone.

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