Integration --Trig functions

[watabi]

Homework Statement

integrate tan^3(9x)

Homework Equations

tan^2(x)= sec^2(x)-1
integral of tanxdx= ln|secx|

The Attempt at a Solution

Integrate tan³(9x)dx
Integral (sec²(x)-1)tan(9x)dx
so then I distribute the tan(9x) giving me:
integral tan(9x)sec²(x)dx - integral tan(9x) dx
so then I solve the first part with u-substitution:
u=tan(9x)
(1/9)du=sec²(9x)
and plug it in the first part
(1/9)integral u - integral tan(9x)
so solving for both
(1/18)tan²(9x)-(1/9)ln |(sec(9x)| +c

(I hope I typed it all correctly)

Any help is appreciated!
[/B]

 

[Chestermiller]

Just try something like you've already done, but only with the sin^3 in the numerator.

Chet

 

[watabi]

Wait so solve sin^3(9x)?
Not sure how that would help but maybe it would go something like:

integral sin³(9x)dx
integral sin²(9x)sin(9x)
we know that sin^2(9x)= 1-cos^2(9x)
so change it up and distribute giving
integral sin(9x)-integral sin(9x)cos²(9x)
which gives me
(-1/9)cos(9x)-integral sin(9x)cos²(9x)
we use u-sub on the second part
u=cos(9x)
(1/9)du=-sin(9x)dx
put it all in giving us
-(1/9)cos(9x)+(1/27)sin^3(9x)+c

Still kinda clueless on the original one though.

 

[Chestermiller]

Wait so solve sin^3(9x)?
Not sure how that would help but maybe it would go something like:

integral sin³(9x)dx
integral sin²(9x)sin(9x)
we know that sin^2(9x)= 1-cos^2(9x)
so change it up and distribute giving
integral sin(9x)-integral sin(9x)cos²(9x)
which gives me
(-1/9)cos(9x)-integral sin(9x)cos²(9x)
we use u-sub on the second part
u=cos(9x)
(1/9)du=-sin(9x)dx
put it all in giving us
-(1/9)cos(9x)+(1/27)sin^3(9x)+c

Still kinda clueless on the original one though.

$$\tan^3θ=\frac{\sin^3θ}{\cos^3θ}=\frac{(1-\cos^2θ)}{\cos^3θ}\sinθ$$