How do you find the integral of int (cotx)^5(sinx)^4dx∫(cotx)5(sinx)4dx?

2 Answers
Sonnhard
Jun 2, 2018

t^4/4+t^2/2-1/2*ln|t+1|-1/2ln|t-1|+Ct44+t22−12⋅ln|t+1|−12ln|t−1|+C where t=cos(x)t=cos(x)

Explanation:

Note that
cot^4(x)*sin^4(x)=cos^5(x)/sin^5(x)*sin^4(x)=cos^5(x)/sin(x)cot4(x)⋅sin4(x)=cos5(x)sin5(x)⋅sin4(x)=cos5(x)sin(x)
Substituting

t=cos(x)t=cos(x)
then we get

dt=-sin(x)dxdt=−sin(x)dx
and our integral will be
-int t^5/(1-t^2)dt−∫t51−t2dt
and this is
-int( -t^3-t-1/(2*(t+1))-1/(2*(t-1)))dt−∫(−t3−t−12⋅(t+1)−12⋅(t−1))dt
this is
-(t^4/4-t^2/2-1/2*ln|t+1|-1/2*ln|t-1|)+C−(t44−t22−12⋅ln|t+1|−12⋅ln|t−1|)+C

maganbhai P.
Jun 2, 2018

I=ln|sinx|-sin^2x+1/4sin^4x+cI=ln|sinx|−sin2x+14sin4x+c

Explanation:

Here,

I=int(cotx)^5(sinx)^4dxI=∫(cotx)5(sinx)4dx

=intcos^5x/sin^5x xxsin^4xdx=∫cos5xsin5x×sin4xdx

=intcos^5x/sinxdx=∫cos5xsinxdx

=int((cos^2x)^2cosx)/sinxdx=∫(cos2x)2cosxsinxdx

=int((1-sin^2x)^2cosx)/sinxdx=∫(1−sin2x)2cosxsinxdx

Subst. sinx=u=>cosxdx=dusinx=u⇒cosxdx=du

So,

I=int(1-u^2)^2/uduI=∫(1−u2)2udu

=int(1-2u^2+u^4)/udu=∫1−2u2+u4udu

=int[1/u-2u+u^3]du=∫[1u−2u+u3]du

=ln|u|-2u^2/2+u^4/4+c=ln|u|−2u22+u44+c

Subst. back .u=sinxu=sinx

I=ln|sinx|-sin^2x+1/4sin^4x+cI=ln|sinx|−sin2x+14sin4x+c

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