Question #6cffa

1 Answer
Jan 31, 2018

The answer is #=3.91#

Explanation:

Apply half angle formulae

#sin(x)=2sin(x/2)cos(x/2)#

#cosx=1-2sin^2(x/2)#

#=>#, #2sin^2(x/2)=1-cosx#

Therefore,

#int(7sin^2xdx)/(sqrt(1-cosx))=7int(4sin^2(x/2)cos^2(x/2)dx)/(sqrt2sin(x/2))#

#=28/sqrt2intsin(x/2)cos^2(x/2)dx#

Apply the substitution

#u=cos(x/2)#, #=>#, #du=-1/2sin(x/2)dx#

Therefore,

#int(7sin^2xdx)/(sqrt(1-cosx))=-28sqrt2intu^2du#

#=-28sqrt2u^3/3#

#=-(28sqrt2)/(3)cos^3(x/2)+C#

Now, calculate the definite integral,

#int_(pi/3)^(pi/2)(7sin^2xdx)/(sqrt(1-cosx))=[-(28sqrt2)/(3)cos^3(x/2)]_(pi/3)^(pi/2)#

#=(-28sqrt2/3*1/(2sqrt2))-(-28sqrt2/3*3sqrt3/8)#

#=(7sqrt3)/(sqrt2)-14/3#

#=3.91#