what is the integral of (sin^(3)x)/cosx?

Question

could you also explain how u got it. thanks so much. i have a test tommorow and keep coming up with the wrong answer.

Answer 1

ok...
break up the sin cubed to sin square times sine

sin^2x*sinx/cos x

now change sin^2x to 1-cos^2x

(sinx)(1-cos^2x)/cos x

which equals (-sinx)(cos^2x-1)/ cos x
then (-sinx) (cos x - sec x)
then -sinxcosx + sinxsecx
so you can split it up into two parts

first part cosx(-sinx), integrate it to get cos^2x/2 + c

second part sinxsecx = -(1/cosx)(-sinx)
as with the rule of integrating 1/x gives ln lxl, thus the integral of the 2nd part is -ln lcosxl + c

thus the answer is cos^2x/2 -ln lcosxl + c

Answer 2

Rewrite (sin x)^3 as (sin x)^2 (sin x)
Then rewrite (sin x)^2 as 1 - (cos x)^2
You have
[ (1 -(cos x)^2)/(cos x)] sin x dx
Now use substitution. Let u = cos x, then du = -sin x dx
Thus, -du = sin x dx
Now the integral becomes
[(1-u^2)/u] (-du)
Divide everything by u and distribute the negative sign.
= (-1/u + u) du
Integrate
= -ln |u| + (1/2)u^2 + C
Replace u with cos x
= -ln|cos x| + (1/2)(cos x)^2 + C