# Evaluating the integral, correct?

by Zack88
Tags: correct, evaluating, integral
 P: 62 lol i was looking at it and then i tilted my head and was like yep i dont know how that happened
 P: 62 woo now i know 6 out of the 18 problems i have to do are correct thanks to you
P: 1,754
 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 :-]
 P: 62 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
 P: 1,754 $$\int\sec^6 xdx$$ $$\int\sec^4 x \sec^2 x dx$$ $$\int(\sec^2 x)^2 \sec^2 xdx$$ When you trig identities raised to powers, break it up till you see something.
 P: 62 [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
P: 1,754
 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!!!
 P: 62 woot lol
 P: 62 im goin to let my brain cool down for the night, t2ul and have a good night.
P: 1,754
 Quote by Zack88 woot lol
Alright, I'm going to sleep! You still have Sunday and Monday to finish all 18 problems!!!
 P: 62 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
 P: 1,754 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
 P: 62 [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
P: 1,754
 Quote by Zack88 [integral] sin^5 x cos^2 x cos x sin^5 x (1- sin^2x) cos x u= sin du= cos [integral] u^5(1- u^2) du u^5 - u^7 du 1/6 sin^6 x - 1/8 sin^8 x + c
Good!
 P: 62 woo, for [integral] x cos^2 x dx do i want to use cos^2 x = (1 + cos^2 x) / 2, then parts or [integral] x (1-sin^2x), then parts?
P: 1,754
 Quote by Zack88 woo, for [integral] x cos^2 x dx do i want to use cos^2 x = (1 + cos^2 x) / 2, then parts or [integral] x (1-sin^2x), then parts?
Try a method! Come test day, you gotta just go at it :-] You've handled harder problems than this, I'm sure you can do this with ease.
 P: 62 sorry had a fire drill and then went to go eat im goin to do the (cos^2 x)/ 2 and then the chain rule
 P: 62 [integral] x cox^2 x dx u= x du = dx dv= cos^2 v= (sin(2x)) / 4 + x/2 x (sin(2x)) / 4 + x/2 - [integral] (sin(2x)) / 4 + x/2 du x (sin(2x)) / 4 + x/2 + cos(2x) / 8 + x^2 / 4 + c

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