intsqrt(1-7w^2)dw∫√1−7w2dw
Consider the substitution:
sqrt(7)w=sintheta√7w=sinθ
-> dw=1/sqrt(7)costheta d theta→dw=1√7cosθdθ
Substituting these into the integral:
intsqrt(1-sin^2theta)*1/sqrt(7)costhetad theta∫√1−sin2θ⋅1√7cosθdθ
We can use 1-sin^2theta=cos^2theta1−sin2θ=cos2θ to get:
=1/sqrt(7)intsqrt(cos^2theta)*costheta d theta=1√7∫√cos2θ⋅cosθdθ
=1/sqrt(7)intcosthetacostheta d theta=1/sqrt(7)intcos^2theta d theta=1√7∫cosθcosθdθ=1√7∫cos2θdθ
Now use the trig identity:
cos^2theta = 1/2+1/2cos2thetacos2θ=12+12cos2θ
So the integral now becomes:
1/sqrt(7)int1/2+1/2cos2theta d theta=1/(2sqrt7)int1+cos2theta d theta1√7∫12+12cos2θdθ=12√7∫1+cos2θdθ
=1/(2sqrt(7))(theta+1/2sin2theta) +C=12√7(θ+12sin2θ)+C
Now use the trig identity:
sin(2theta)=2costhetasinthetasin(2θ)=2cosθsinθ to get:
=1/(2sqrt(7))(theta+costhetasintheta) +C=12√7(θ+cosθsinθ)+C
=1/(2sqrt(7))(theta+sqrt(1-sin^2theta)sintheta) +C=12√7(θ+√1−sin2θsinθ)+C
Now we can reverse the substitution we started with:
=1/(2sqrt(7))(sin^(-1)(sqrt7w)+sqrt(1-7w^2)sqrt7w) +C=12√7(sin−1(√7w)+√1−7w2√7w)+C
=1/(2sqrt(7))sin^(-1)(sqrt7w)+w/2sqrt(1-7w^2) +C=12√7sin−1(√7w)+w2√1−7w2+C