Question

integrationn of cos2xsinx

Answer 1

cosA sinB = 1/2(sin(A+B) - sin(A-B))

Answer 2

One way would be ...
write cos2x= 1-2 sin^2x ....

Answer 3

int (cos2x sinx) dx
= int (1/2(sin(2x+x) - sin(2x-x))
= 1/2 int (sin3x - sinx) dx
= 1/2 int sin 3x dx - 1/2 int sinx dx

does that helps ?

Answer 4

better thing might be to write cos2x = 2cos^2x - 1 and assume cos^2x =t

Answer 5

I mean assume cosx = t

Answer 6

hint : use the basic formulas
int sinx = -cosx and
int sinAx = -1/A cosAx

Answer 7

thank you :) it is helpful:D

Answer 8

can you give me the final answer? i want to see if i correct:/

Answer 9

well, actually that's just a basic integration problem :)
1/2 int sin 3x dx - 1/2 int sinx dx
= 1/2(-1/3 cos3x) - 1/2(-cosx) + c
= -1/6 cos3x + 1/2 cosx + c

Answer 10

I just found the MS, the answer is \[-2/3\cos ^{3}x+cosx\]

Answer 11

here is detailed solution

Answer 12

thank you ~ maybe the MS is wrong:D