# Trig substitution

1. Jun 9, 2006

### suspenc3

How would I go about solving this:

$$\int cos^2x tan^3xdx$$..all i did so far ..

$$\int cos^2x tanx(tan^2x)$$

$$[[ \frac{1}{cosx}^2 -1] = tan^2x$$

so...

$$\int cos^2x[[ \frac{1}{cosx}]^2-1]tanx$$

is this right so far...now what?

Last edited: Jun 9, 2006
2. Jun 9, 2006

### TD

Another way would be using tan(x) = sin(x)/cos(x), like:

$$\cos ^2 x\tan ^3 x = \cos ^2 x\frac{{\sin ^3 x}} {{\cos ^3 x}} = \frac{{\sin ^3 x}} {{\cos x}}$$

Now if you use sin³x = sin²x.sinx and convert the sin²x to 1-cos²x, I smell a good substitution coming

Your method was fine as well though, continuing could give:

$$\cos ^2 x\tan ^3 x = \cos ^2 x\left( {\sec ^2 x - 1} \right)\tan x = \tan x - \cos ^2 x\tan x = \tan x - \sin x\cos x$$

Last edited: Jun 9, 2006
3. Jun 9, 2006

### suspenc3

ohh..ok so i got as far as $$\int \frac{sinx - sinxcos^2x}{cosx}$$

4. Jun 9, 2006

### TD

Well, you can do many things with that. My choice would be rewriting as:

$$\int {\frac{{\sin x - \sin x\cos ^2 x}} {{\cos x}}dx} = \int {\frac{{\sin x\left( {1 - \cos ^2 x} \right)}} {{\cos x}}dx}$$

Do you see a nice substitution now?

5. Jun 9, 2006

### suspenc3

yes I get $$ln|cosx| - \frac{1}{2}cos^2x + C$$

6. Jun 9, 2006

### TD

You should check that again (signs?)

7. Jun 9, 2006

### suspenc3

im having trouble with another..ive worked it down..but i think its wrong..

$$\int cot^5xsin^4xdx$$

$$\int \frac{cos^5x}{sinx}dx$$...would i just simplify the top to cos^2x(cos^2x)(cosx)?

8. Jun 9, 2006

### TD

Sure, and switch each cos²x to 1-sin²x so you can substitute y = sin(x).

Watch out though: your last one wasn't correct. I find:

$$- \ln \left( {\cos x} \right) - \frac{{\sin ^2 x}} {2} + C$$

9. Jun 9, 2006

### suspenc3

yea i messed up the signs thanks|

10. Jun 9, 2006

### TD

Ok, the other one should be fine now?

$$\int {\frac{{\cos ^5 x}} {{\sin x}}dx} = \int {\frac{{\left( {1 - \sin ^2 x} \right)^2 }} {{\sin x}}\cos xdx}$$

Perfect for y = sin(x).

11. Jun 9, 2006

### suspenc3

how about $$\int sec^6tdt$$

where to start?

sec^2t(sec^2t)(sec^2t)?

Last edited: Jun 9, 2006
12. Jun 9, 2006

### motai

The first thing you want to do is split up $$\int sec^6tdt$$ into components so some parts of it will cancel.

You're on the right track, remember that sec^2t = 1 + tan^2t, and make a u substitution. If you go about it correctly, one of those sec^2ts will cancel and you will have the integral in terms of u. Once you get to that step, backsubstitute and tack on the C at the end.

Best of luck.

13. Jun 12, 2006

### VietDao29

Uhm, sec6x = sec6x - sec4x + sec4x - sec2x + sec2x = sec4x (sec2x - 1) + sec2x (sec2x - 1) + sec2x.
And you should also notice that:
$$\frac{d}{dx} \tan x = \sec ^ 2 x$$
I'll give you an example.
---------------
Example:
$$\int \sec ^ 4 x dx = \int \left( \sec ^ 4 x - \sec ^ 2 x + \sec ^ 2 x \right) dx = \int \left( \sec ^ 2 x (\sec ^ 2 x - 1) + \sec ^ 2 x \right) dx$$
$$= \int ( \sec ^ 2 x \tan ^ 2 x ) dx + \int \sec ^ 2 x dx = \int \tan ^ 2 x d(\tan x) + \int \sec ^ 2 x dx = \frac{\tan ^ 3 x}{3} + \tan x + C$$
Can you go from here? :)