integrating with trigonometric functions

[mooshupork34]

hi, I was wondering if someone could show me how to do a problem. I'm studying for a test...and I wanted to be prepared in case something like this came up. It confused me quite a bit...

how to you integrate the following...

(cosine cubed of x) * (sin to the fifth of x) dx

 

[soroban]

Hello, mooshupork34!

\(\displaystyle \L\int\cos^3x\cdot\sin^5x\,dx\)

We have: \(\displaystyle \:\cos^3x\cdot\sin^5x\;=\;\cos^2x\cdot\sin^5x\cdot\cos x\)

. . . \(\displaystyle = \;(1\,-\,sin^2x)(\sin^5x)(\cos x)\;=\;(\sin^5x\,-\,\sin^7x)(\cos x)\)


The integral becomes: \(\displaystyle \L\:\int(\sin^5x\,-\,\sin^7x)(\cos x\,dx)\)

Substitute: \(\displaystyle u\,=\,\sin x\) . . . and get: \(\displaystyle \:\int(u^5\,-\,u^7)\,du\)

Got it?

 

[mooshupork34]

thanks! so here's what i did...

i set u = sin(x), which means that du/dx = cos(x).

then, this means that dx = 1/cos(x) * du

then that means that you would have to take the integral of

(u^5 - u^7) * (1/cos(x) dx)

this is where i got stuck...

 

[Steem]

Don't forget you only substituted for sinx, this still leaves you with (u^5-u^7)(cosx)(1/cosx dx)
Cancel out and the rest is a breeze!

 

[mooshupork34]

WOW. Man, that was stupid of me.

Thank you so much, everyone!