Question #d2780

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Answer 1 · 2017-11-06T12:07:03

# 1/(2sqrt3)tan^-1(1/sqrt3*tan2x)+C.#

Part I :

Let, #I=int1/(2cos^2 2x+1)dx,#

#=int1/{cos^2 2x(2+1/cos^2(2x))}dx,#

#=int(1/cos^2 (2x))/(2+sec^2 2x)dx,#

#=int(sec^2 2x)/{2+(1+tan^2 2x)}dx,#

# :. I=int(sec^2 2x)/(3+tan^2 2x)dx.#

We subst. #tan 2x=y," so that, "(sec^2 2x)(2)dx=dy.#

#:. I=int((1/2)dy)/(3+y^2),#

#=1/2*1/sqrt3*tan^-1(y/sqrt3),#

# rArr I=1/(2sqrt3)tan^-1(1/sqrt3*tan2x)+C.#

Part II :

It is not clear that what is to be integrated.