Trigonometric Integration

[C.C.]

I really need help with this integration problem.

I have so far:

(1-sinx^2)^2 * cosx * tanx^2 * tan x)

But how do I continue?

 

[Deleted member 4993]

I really need help with this integration problem.

I have so far:

(1-sinx^2)^2 * cosx * tanx^2 * tan x)

But how do I continue?

Use

\(\displaystyle tan^3(x) = \frac{sin^3(x)}{cos^3(x)}\)

 

[C.C.]

Alright, I used that and got:

INTEGRAL (1-(sinx)^2)^2 * cosx * (sinx)^3/(cosx)^3

(du/cosx) = dx

INTEGRAL (1-u^2)^2 * (u^3/(cosx)^3)

(1-u^2)^2 == 1

INTEGRAL 1 * u^3 * (1/(cosx)^3)

Is this right so far?

 

[BigGlenntheHeavy]

\(\displaystyle \int cos^{5}(x)tan^{3}(x)dx \ = \ \int cos^{5}(x)\frac{sin^{3}(x)}{cos^{3}(x)}dx \ = \ \int cos^{2}(x)sin^{3}(x)dx.\)

\(\displaystyle = \ \int cos^{2}(x)sin^{2}(x)sin(x)dx \ = \int cos^{2}(x)sin(x)[1-cos^{2}(x)]dx\)

\(\displaystyle = \ \int [cos^{2}(x)sin(x)-cos^{4}(x)sin(x)]dx \ = \ \int cos^{2}(x)sin(x)dx-\int cos^{4}(x)sin(x)dx\)

\(\displaystyle You \ should \ be \ able \ to \ take \ it \ from \ here.\)