Asked by Rianne
Evaluate the Integral
pi
/ = (3cos^2x-1) (sinx) dx=0
0
I am really confused with this, please help!
** used / for integral, not sure if there is a way to type it!
pi
/ = (3cos^2x-1) (sinx) dx=0
0
I am really confused with this, please help!
** used / for integral, not sure if there is a way to type it!
Answers
Answered by
Steve
∫[0,pi] (3cos^2x-1) (sinx) dx
let u = cosx, then du = -sinx dx
now you have
∫[1,-1] -(3u^2-1) du
= u^3 - u [-1,1]
= 0
let u = cosx, then du = -sinx dx
now you have
∫[1,-1] -(3u^2-1) du
= u^3 - u [-1,1]
= 0
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.