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