## Evaluating the integral, correct?

$$\int \cos^{3}x\sin x dx$$

- cos^4 x / 4

 Quote by Zack88 -4sin^3?
 Quote by Zack88 $$\int \cos^{3}x\sin x dx$$ - cos^4 x / 4
Yep. Now look at this problem, but don't pay attention the first part, the 2nd step is what I'm mainly talking about. I spent like an hour evaluating it through Parts a couple times, till I got tired and asked for helped and look how simple it was ...

 I see how you got to step 3 but not to 4

 Quote by Zack88 I see how you got to step 3 but not to 4
Oh, don't worry about that. I only wanted to emphasize the chain-rule. Step 3 to 4 would take me forever to type, lol.

I simplified a lot to save myself typing-time. I basically used a trig identity.

 lol i was looking at it and then i tilted my head and was like yep i dont know how that happened
 woo now i know 6 out of the 18 problems i have to do are correct thanks to you

 Quote by Zack88 woo now i know 6 out of the 18 problems i have to do are correct thanks to you
I'm getting sleepy, if you wanna do a couple more better start asking :-]

 me too me too [integral] sec^6 x dx should i do [integral] (sec^2)^3 or start doing parts u= sec^2 x dv = sec^4 x
 $$\int\sec^6 xdx$$ $$\int\sec^4 x \sec^2 x dx$$ $$\int(\sec^2 x)^2 \sec^2 xdx$$ When you trig identities raised to powers, break it up till you see something.
 [integral] (tan^2 x +1) sec^2 x dx u = tan x du = sec^2 x [integral] (u^2 =1)^2 du [integral]u^4 + 2u^2 + 1 du tan^5 x / 5 + 2tan^3 x / 3 + tan x + c

 Quote by Zack88 [integral] (tan^2 x +1) sec^2 x dx u = tan x du = sec^2 x [integral] (u^2 =1)^2 du [integral]u^4 + 2u^2 + 1 du tan^5 x / 5 + 2tan^3 x / 3 + tan x + c
Good!!!

 woot lol
 im goin to let my brain cool down for the night, t2ul and have a good night.

 Quote by Zack88 woot lol
Alright, I'm going to sleep! You still have Sunday and Monday to finish all 18 problems!!!

 woo im back, v.v [integral] sin^5 x cos^3 x i was wanting to know if i should break it up like [integral] sin^4 x sin x cos^2 x cos x
 Look in your book, there should be a suggestion on how to tackle even/odd powers of sines and cosines. Hint: Leave sin^5 x alone, mess around with cos^3 x
 [integral] sin^5 x cos^2 x cos x [integral] sin^5 x (1- sin^2x) cos x u= sin du= cos [integral] u^5(1- u^2) du [integral] u^5 - u^7 du 1/6 sin^6 x - 1/8 sin^8 x + c