Question

integral of sinx sinx(cosx) dx

Answer 1

substitute u = cosx, then du = -sinxdx

so it becomes the integral of -sin(u)du

= cos(u) + C

= cos(cosx) + C

Answer 2

oh.. but can i also use \[\int\limits_{}^{}\sin x dx \int\limits_{}^{}six(cosx)dx\]

Answer 3

what is six? a typo or a new function i haven't heard of?

Answer 4

sorry for the typo. that's sin. xD

Answer 5

it is not necessarily true that sin(x) = sin(cosx)

Answer 6

aw.. the given is \[\int\limits_{}^{} \sin x \sin(cosx) dx\]

Answer 7

yes...and that integral is cos(cosx) + C

are you aware of the chain rule?

Answer 8

yes. never thought of that. ty btw. :)

Answer 9

well, use the chain rule to differentiate cos(cosx) + C

what do you get?

Answer 10

hbaldon, you should follow jamesm method. You can not break it up the way you did because integral has restriction on multiplication; you can break it up like that if it has a + or - between them.