Question #62823

1 Answer
Jan 7, 2018

=1/(2sqrt(7))sin^(-1)(sqrt7w)+w/2sqrt(1-7w^2) +C=127sin1(7w)+w217w2+C

Explanation:

intsqrt(1-7w^2)dw17w2dw

Consider the substitution:

sqrt(7)w=sintheta7w=sinθ

-> dw=1/sqrt(7)costheta d thetadw=17cosθdθ

Substituting these into the integral:

intsqrt(1-sin^2theta)*1/sqrt(7)costhetad theta1sin2θ17cosθdθ

We can use 1-sin^2theta=cos^2theta1sin2θ=cos2θ to get:

=1/sqrt(7)intsqrt(cos^2theta)*costheta d theta=17cos2θcosθdθ

=1/sqrt(7)intcosthetacostheta d theta=1/sqrt(7)intcos^2theta d theta=17cosθcosθdθ=17cos2θdθ

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 theta1712+12cos2θdθ=1271+cos2θdθ

=1/(2sqrt(7))(theta+1/2sin2theta) +C=127(θ+12sin2θ)+C

Now use the trig identity:

sin(2theta)=2costhetasinthetasin(2θ)=2cosθsinθ to get:

=1/(2sqrt(7))(theta+costhetasintheta) +C=127(θ+cosθsinθ)+C

=1/(2sqrt(7))(theta+sqrt(1-sin^2theta)sintheta) +C=127(θ+1sin2θsinθ)+C

Now we can reverse the substitution we started with:

=1/(2sqrt(7))(sin^(-1)(sqrt7w)+sqrt(1-7w^2)sqrt7w) +C=127(sin1(7w)+17w27w)+C

=1/(2sqrt(7))sin^(-1)(sqrt7w)+w/2sqrt(1-7w^2) +C=127sin1(7w)+w217w2+C