Question

24. If I = [sec? 2xdx , then the value of 1 is 4 1) tan x + c 2) tan 2x + c 5 tan 2x 3) + c 4) cot 2 x + c 2​

Answer 1

Answer:

A

K=−1/3

C

M=−2

Let , I=∫sec

2

xcsc

4

xdx

Using integration by parts ,

I=csc

4

x∫sec

2

xdx−∫

dx

d

(csc

4

x)∫sec

2

xdx

=csc

4

xtanx−∫(−4csc

4

xcotx)tanxdx

=csc

4

xtanx+4∫csc

4

xdx

=csc

4

xtanx+4∫csc

2

xcsc

2

xdx

applying integration by parts again.

=csc

4

xtanx+4[csc

2

x∫csc

2

xdx−∫(−2csc

2

xcotx)(−cotx)dx]

=csc

4

xtanx+4[csc

2

x(−cotx)−2∫csc

2

xcot

2

xdx]

=csc

4

xtanx−4csc

2

xcotx+

3

8

cot

3

x

Substitute , csc

2

x=1+cot

2

x and simplify

Then we get ,

I=−

3

1

cot

3

x+tanx−2cotx+C

Therefore , K=−

3

1

L=1 M=−2