Question

∫tan32x sec2x dx = __________.​

Answer 1

Step-by-step explanation:

LetI=∫tan

3

2x.sec2xdx

=∫tan

2

2x.sec2x.tan2xdx

=∫(sec

2

2x−1)sec2x.tan2xdx

Put sec2x=t

2sec2x.tan2xdx=dt

Now sec2x.tan2xdx=

2

dt

I=

2

1

∫(t

2

−1)dt

=

2

1

[

3

t

3

−t]+C

=

2

1

[

3

sec

3

2x

−sec2x]+C