Integration with Trig (Calc 2)

[demersal]

Homework Statement

\int\frac{(sin x)^{3}}{(cos x)}dx

Homework Equations

Trigonometric identities

The Attempt at a Solution

\int\frac{(sin x)^{2}}{(cos x)}(sin x)dx

\int\frac{1-(cos x)^{2}}{(cos x)}(sin x)dx

u = cos x
du = -sin x dx

- \int\frac{1-(u)^{2}}{(u)}du

Where do I go from here? I am kind of stuck. Can I just simply split the two up, as in 1/u - u^2/2? I think I tried that and end up with the wrong answer. Thanks in advance for the advice!

 

[Dick]

Sure you can split them up. But what happened to the '-' in -sin(x)dx?

 

[demersal]

Oops! I had the - on my paper, I forgot it in transferring it I will edit that now!

But the answer I am getting is not equal to the answers those online integral calculators are showing. Here is my work:

= - \intdu/u + \int\frac{u^{2}}{u}du
= - ln u + (1/2)u^{2} + c
= - ln (cos x) + ((cos x)^2)/2 + c

Am I making some sort of elementary mistake?

 

[Dick]

Oops! I had the - on my paper, I forgot it in transferring it I will edit that now!

But the answer I am getting is not equal to the answers those online integral calculators are showing. Here is my work:

= - \intdu/u + \int\frac{u^{2}}{u}du
= - ln u + (1/2)u^{2} + c
= - ln (cos x) + ((cos x)^2)/2 + c

Am I making some sort of elementary mistake?

That's perfectly fine. Differentiate it to make sure. Now take one on the online integrator solutions and do the same. There's lots of different ways to write the result that only differ by a constant. For example instead of cos(x)^2/2 you could put -sin(x)^2/2. They only differ by a constant.

 

[demersal]

Thanks, Dick! It did work. I feel very silly for not thinking of doing that myself!