Integrate (sinx+cosx)/√(sin2x)?

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Answer 1 · 2017-09-10T17:27:15

#arc sin(sinx-cosx)+C, or, arc sin{sqrt2sin(x-pi/4)}+C.#

Let, #I=int(sinx+cosx)/sqrt(sin2x)dx.#

We subst., #sinx-cosx=t rArr (cosx+sinx)dx=dt.#

Further,

#t^2=(sinx-cosx)^2=sin^2x-2sinxcosx+cos^2x, i.e., #

# t^2=1-sin2x rArr sin2x=1-t^2.#

#:. I=intdt/sqrt(1-t^2),#

#=arc sint,#

# rArr I=arc sin(sinx-cosx)+C.#

Since, #sinx-cosx=sqrt2(1/sqrt2sinx-1/sqrt2cosx),#

#=sqrt2{sinxcos(pi/4)-cosxsin(pi/4)},#

#=sqrt2sin(x-pi/4),# we can write,

#I=arc sin{sqrt2sin(x-pi/4)}+C.#

Enjoy Maths!