Question #62823

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Answer 1 · 2018-01-07T14:00:29

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

$intsqrt(1-7w^2)dw$

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:

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

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