What is int (sin^2x)/(cos^3(x)) dx∫sin2xcos3(x)dx?

2 Answers
mason m
Jun 18, 2018

1/2secxtanx-1/2lnabs(secx+tanx)+C12secxtanx−12ln|secx+tanx|+C

Explanation:

intsin^2x/cos^3xdx=int(1-cos^2x)/cos^3xdx=int(sec^3x-secx)dx∫sin2xcos3xdx=∫1−cos2xcos3xdx=∫(sec3x−secx)dx

Both of these are fairly tricky. Here's a link that should help out a lot: https://www.math.ubc.ca/~feldman/m121/secx.pdf

Using these results, we get an answer:

=(1/2secxtanx+1/2lnabs(secx+tanx))-lnabs(secx+tanx)+C=(12secxtanx+12ln|secx+tanx|)−ln|secx+tanx|+C

=1/2secxtanx-1/2lnabs(secx+tanx)+C=12secxtanx−12ln|secx+tanx|+C

Ratnaker Mehta
Jun 18, 2018

1/2[secxtanx-ln|(secx+tanx)|+C12[secxtanx−ln|(secx+tanx)|+C,

or, 1/2[sinx/cos^2x-ln|(1+sinx)/cosx|]+Cor,12[sinxcos2x−ln∣∣∣1+sinxcosx∣∣∣]+C.

Explanation:

Let, I=intsin^2x/cos^3xdx=intsinx/cosx*1/cosx*sinx/cosxdxI=∫sin2xcos3xdx=∫sinxcosx⋅1cosx⋅sinxcosxdx,

=int(tanx)(secxtanx)dx=∫(tanx)(secxtanx)dx.

So, letting, secx=t," we have, "secxtanxdx=dtsecx=t, we have, secxtanxdx=dt.

Also, tanx=sqrt(sec^2x-1)tanx=√sec2x−1.

:. I=intsqrt(t^2-1)dt,

=1/2[tsqrt(t^2-1)-ln|(t+sqrt(t^2-1))|].

Since, t=secx, we get,

I=1/2[secxtanx-ln|(secx+tanx)|+C,

or, I=1/2[sinx/cos^2x-ln|(1+sinx)/cosx|]+C.

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