#intsqrt(1-7w^2)dw#
Consider the substitution:
#sqrt(7)w=sintheta#
#-> dw=1/sqrt(7)costheta d theta#
Substituting these into the integral:
#intsqrt(1-sin^2theta)*1/sqrt(7)costhetad theta#
We can use #1-sin^2theta=cos^2theta# to get:
#=1/sqrt(7)intsqrt(cos^2theta)*costheta d theta#
#=1/sqrt(7)intcosthetacostheta d theta=1/sqrt(7)intcos^2theta d theta#
Now use the trig identity:
#cos^2theta = 1/2+1/2cos2theta#
So the integral now becomes:
#1/sqrt(7)int1/2+1/2cos2theta d theta=1/(2sqrt7)int1+cos2theta d theta#
#=1/(2sqrt(7))(theta+1/2sin2theta) +C#
Now use the trig identity:
#sin(2theta)=2costhetasintheta# to get:
#=1/(2sqrt(7))(theta+costhetasintheta) +C#
#=1/(2sqrt(7))(theta+sqrt(1-sin^2theta)sintheta) +C#
Now we can reverse the substitution we started with:
#=1/(2sqrt(7))(sin^(-1)(sqrt7w)+sqrt(1-7w^2)sqrt7w) +C#
#=1/(2sqrt(7))sin^(-1)(sqrt7w)+w/2sqrt(1-7w^2) +C#
