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

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Answer 1 · 2018-06-18T19:30:16

#1/2secxtanx-1/2lnabs(secx+tanx)+C#

#intsin^2x/cos^3xdx=int(1-cos^2x)/cos^3xdx=int(sec^3x-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#

#=1/2secxtanx-1/2lnabs(secx+tanx)+C#

Answer 2 · 2018-06-18T19:36:18

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

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

Let, #I=intsin^2x/cos^3xdx=intsinx/cosx*1/cosx*sinx/cosxdx#,

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

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

Also, #tanx=sqrt(sec^2x-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#.