Question

\[Evaluate ==> \sin ^{2}xcosxdx\]

Answer 1

do you mean integral?

Answer 2

You can write:

\[\sin^2xcosx = (1-\cos^2x).(cosx) = cosx - \cos^3x\]

So.

\[\int\limits cosx.dx - \int\limits \cos^2x dx\]

Answer 3

Try to solve this..

Answer 4

careful

Answer 5

You are scaring me..

Answer 6

you can just simply do \[\int \sin^2 x \cos x dx\]
let u = \(\sin x)\)
du = \(\cos x\)

\[\int \sin^2 x \cos x dx \implies \int u^2 du\]
now it's just power rule

Answer 7

no need to do the complex integral such as cos^2 x dx

Answer 8

oh yeah..

Answer 9

wait...how do you integrate cos^3 dx anyway o.O

Answer 10

you'll have to go back to sin^2 x cos x lol

Answer 11

haha. :D

Answer 12

NO..

Answer 13

\[\cos^3x = \frac{\cos(3x) + 3\cos(x)}{4}\]