Question

integration of sin ^{5}X

Answer 1

you want answer in terms of cos3x, cos 5x or in terms of cos^3x , cos^5 x ?

Answer 2

in terms of cos^3x , cos^5 x please:)

Answer 3

well, if you want it in terms of cos^3 x and cos^5 x
what you can do is,
sin^5x = (sin^2 x)^2 . sin x
and then write sin^2x as 1-cos^2x
to get
\(\large \int (1-\cos^2x)^2\sin x dx\)
now you can put u= cos x
du=.... ?

Answer 4

yea good gohan :)

Answer 5

thanks :)

Answer 6

tell me if you have any doubts solving further.

Answer 7

is there way of not using integration by substitution?

Answer 8

if you can express sin^5 x in terms of first powers of sin 3x and sin 5x, then without substitution, you can get the answer, but in terms of cos 3x and cos 5x

Answer 9

the answer is -cosx +2/3 cosx^3x-1/5cos^5x... but I don't know how the substitution works:/

Answer 10

did you get what i explained ?
did you try u=cos x ? du=..... ?
if yes, where are u stuck ?

Answer 11

is it du=-sinx dx?

Answer 12

yes,
so the integral turns to be
\(\large -\int (1-u^2)du\)
right ?

Answer 13

***
\(\large -\int (1-u^2)^2du\)

Answer 14

oh right!! thank you:))

Answer 15

welcome ^_^