Question

How would you solve...

Answer 1

\[\int\limits_{}^{}\frac{ \sin ^3x }{ \sqrt{cosx} }dx\]

Answer 2

@ParthKohli

Answer 3

put t= cos x
dt = sin x dx
split sin^3 x as sin x sin^2 x
write sin^2 x as 1-cos^2 x

Answer 4

so you get 1-t^2/ sqrt u

Answer 5

hi, lalaly :)

Answer 6

*****(1-t^2)/ sqrt t

Answer 7

@Kikazo , tried that ?

Answer 8

Yeah, and you're right! I think your way is easier than mine, I was doing this:



So that \[cosx=u^2\]\[=> -sinx dx=2u du => dx=\frac{ -2udu }{ sinx } => dx=\frac{ -2udu }{ \sqrt{1-u^4} }\]

Then, \[-\int\limits_{}^{}\frac{ (1-u^4)^{3/2} *\frac{ 2udu }{ \sqrt{1-u^4} }}{u}=-2\int\limits_{}^{}(1-u^4)du\], which ends up being the same as in your way