Question

Very hard indefinite integral
cosx*(sinx)^(1/2)

Answer 1

try u=sinx

Answer 2

So u = cos x and dv = sin^0.5
du = -sin x and v, here is my uncertainty is v = -cos^(1/2)?
if so then u'v+uv' = -sinx*sinx^1/2 +cosx *-cosx ^1/2

Answer 3

where do u get sin x from?

Answer 4

by parts not necessary, or even possible here
simple u sub
I got the sinx from inside the (sinx)^1/2

Answer 5

\[\int\limits_{}^{} \cos x * \sqrt{sinx}\]

Answer 6

indeed, let u=sinx
then du=....

Answer 7

okey du is then cos x

Answer 8

so the integral is then what? (in terms of u)

Answer 9

integral of u*dv ?

Answer 10

why do you insist on trying to integrate by parts? there is no dv

Answer 11

or u'v+uv' = (uv)'

Answer 12

oh sorry

Answer 13

sin x * -cos x?

Answer 14

\[\int\sqrt{\sin x}\cos xdx\\u=\sin x\\du=\cos xdx\]rewrite in terms of \(u\)