Solving Trig Substitution: \int cos^2x tan^3xdx

[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?

 

[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:

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

 

[suspenc3]

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

 

[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?

 

[suspenc3]

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

 

[TD]

You should check that again (signs?)

 

[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)?

 

[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:

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

 

[suspenc3]

yea i messed up the signs thanks|

 

[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).

 

[suspenc3]

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

where to start?

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

 

[motai]

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

where to start?

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

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.

 

[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? :)