Trigonometric Integration

C.C.

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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?
 
C.C. said:
0dde99d2413ac934ea7a3d852aca551.png


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

tan3(x)=sin3(x)cos3(x)\displaystyle tan^3(x) = \frac{sin^3(x)}{cos^3(x)}
 
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?
 
cos5(x)tan3(x)dx = cos5(x)sin3(x)cos3(x)dx = cos2(x)sin3(x)dx.\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.

= cos2(x)sin2(x)sin(x)dx =cos2(x)sin(x)[1cos2(x)]dx\displaystyle = \ \int cos^{2}(x)sin^{2}(x)sin(x)dx \ = \int cos^{2}(x)sin(x)[1-cos^{2}(x)]dx

= [cos2(x)sin(x)cos4(x)sin(x)]dx = cos2(x)sin(x)dxcos4(x)sin(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

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