Question #62823

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Question #62823

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Answer 1 · 2018-01-07T14:00:29

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

Explanation:

$\int \sqrt{1 - 7 w^{2}} d w$ $intsqrt(1-7w^2)dw$

Consider the substitution:

$\sqrt{7} w = sin θ$ $sqrt(7)w=sintheta$

$\to d w = \frac{1}{\sqrt{7}} cos θ d θ$ $-> dw=1/sqrt(7)costheta d theta$

Substituting these into the integral:

$\int \sqrt{1 - {sin}^{2} θ} \cdot \frac{1}{\sqrt{7}} cos θ d θ$ $intsqrt(1-sin^2theta)*1/sqrt(7)costhetad theta$

We can use $1 - {sin}^{2} θ = {cos}^{2} θ$ $1-sin^2theta=cos^2theta$ to get:

$= \frac{1}{\sqrt{7}} \int \sqrt{{cos}^{2} θ} \cdot cos θ d θ$ $=1/sqrt(7)intsqrt(cos^2theta)*costheta d theta$

$= \frac{1}{\sqrt{7}} \int cos θ cos θ d θ = \frac{1}{\sqrt{7}} \int {cos}^{2} θ d θ$ $=1/sqrt(7)intcosthetacostheta d theta=1/sqrt(7)intcos^2theta d theta$

Now use the trig identity:

${cos}^{2} θ = \frac{1}{2} + \frac{1}{2} cos 2 θ$ $cos^2theta = 1/2+1/2cos2theta$

So the integral now becomes:

$\frac{1}{\sqrt{7}} \int \frac{1}{2} + \frac{1}{2} cos 2 θ d θ = \frac{1}{2 \sqrt{7}} \int 1 + cos 2 θ d θ$ $1/sqrt(7)int1/2+1/2cos2theta d theta=1/(2sqrt7)int1+cos2theta d theta$

$= \frac{1}{2 \sqrt{7}} (θ + \frac{1}{2} sin 2 θ) + C$ $=1/(2sqrt(7))(theta+1/2sin2theta) +C$

Now use the trig identity:

$sin (2 θ) = 2 cos θ sin θ$ $sin(2theta)=2costhetasintheta$ to get:

$= \frac{1}{2 \sqrt{7}} (θ + cos θ sin θ) + C$ $=1/(2sqrt(7))(theta+costhetasintheta) +C$

$= \frac{1}{2 \sqrt{7}} (θ + \sqrt{1 - {sin}^{2} θ} sin θ) + C$ $=1/(2sqrt(7))(theta+sqrt(1-sin^2theta)sintheta) +C$

Now we can reverse the substitution we started with:

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

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

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Question #62823

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