# Thread: integration problem

1. ## integration problem

I have been doing couple of integration by parts and when I got to trig. integrals I get confuse, specially this problem.

Integral cos x^2 tan x^3 dx

I'm used to problems that has tan with sec^2's, cos with cos^2's etc
but not different trig functions such as cos and tan.

2. Originally Posted by ff4930
I have been doing couple of integration by parts and when I got to trig. integrals I get confuse, specially this problem.

Integral cos x^2 tan x^3 dx

I'm used to problems that has tan with sec^2's, cos with cos^2's etc
but not different trig functions such as cos and tan.
i'm pretty sure that $\displaystyle \cos \left( x^2 \right) \tan \left( x^3 \right)$ is not integrable in terms of elementary functions

did you mean $\displaystyle \int \cos^2 x \tan^3 x~dx$ ?? those are two very different things. be specific as to what you want, and careful

3. Yes, I apologize it is cos^2x and tan^3x.

4. $\displaystyle \cos^2 x \tan^3 x = \cos ^2 x \cdot \frac{\sin^3 x}{\cos ^3 x} = \frac{(1-\cos^2 x)\sin x}{\cos x}$

Let $\displaystyle t=\cos x$ and make subsitution.

5. Originally Posted by ff4930
Yes, I apologize it is cos^2x and tan^3x.
Let me practice on integration.

INT.[cos^2(X) tan^3(X)]dX
= INT.[cos^2(X) sin^3(X)/cos^3(x)]dX
= INT.[ sin^3(X) /cosX]dX
= INT.[{1 -cos^2(X)}sinX/cosX]dX
= INT.[sinX/cosX -sinXcosX]dX
= INT.[tanX -(1/2)sin(2X)]dX
= INT.[tanX]dX -(1/2)INT.[sin(2X)]dX
= ln[secX] -(1/2)(1/2)INT.[sin(2X)](2dX)
= ln[secX] -(1/4)[-cos(2X)]
= ln[secX] +(1/4)cos(2X) +C --------------answer.