Integrate:∫(tan32xsec2x)dx
Consider the given integral.
I=∫(tan32xsec2x)dx
I=∫(tan22xtan2xsec2x)dx
I=∫((sec22x−1)tan2xsec2x)dx
Let t=sec2x
dtdx=sec2xtan2x(2)
dt2=sec2xtan2xdx
Therefore,
I=12∫(t2−1)dt
I=12[t33−t]+C
On putting the value of t, we get
I=12[sec32x3−sec2x]+C
Hence, this is the answer.