Question

pls answer as fast as possible​

Answer 1

Answer:

Correct option is

A

x

2

[−

xtanx+1

1

]+2log(xsinx+cosx)+C

I=∫

(xtanx+1)

2

x

2

(xsec

2

x+tanx)

dx

Note that d.c. of xtanx+1 is xsec

2

x+tanx.

Hence integrating by parts, we get

x

2

[−

xtanx+1

1

]+2∫

xtanx+1

x

dx

Now write I

1

=2∫

xsinx+cosx

xcosx

=2log(xsinx+cosx)

∵ d.c. of xsinx+cosx=xcosx

Thus I=x

2

[−

xtanx+1

1

]+2log(xsinx+cosx)+C