Question #62823

1 Answer
Jan 7, 2018

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

Explanation:

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