How to solve the integration #int sin^2x /(sqrt(1-cosx) )dx# ?

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Answer 1 · 2018-06-25T06:05:01

#intsin^2x/sqrt(1-cosx)dx=(-4sqrt2)/3cos^3(x/2)+c#

Here,

#I=intsin^2x/sqrt(1-cosx)dx=int(sin^2x)/sqrt(1- cosx)xxsqrt(1+cosx)/sqrt(1+cosx)dx#

#I=int(sin^2xsqrt(1+cosx))/sqrt(1-cos^2x)dx#

#=int(sin^2xsqrt(1+cosx))/sinx dx#

#=intsinxsqrt(2cos^2(x/2))dx#

#=int2sin(x/2)cos(x/2)*sqrt2cos(x/2)dx#

#=2sqrt2intcos^2(x/2)sin(x/2) dx#

#=2sqrt2int[cos(x/2)]^2(-1/2sin(x/2))(-2)dx#

#=-4sqrt2int[cos(x/2)]^2d/(dx)(cos(x/2))dx#

#=-4sqrt2[cos(x/2)]^3/3+c#

#=(-4sqrt2)/3cos^3(x/2)+c#