How do you integrate $\int x^{2} \sqrt{16 - x^{2}}$ $int x^2sqrt(16-x^2)$ using trig substitutions?

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How do you integrate $\int x^{2} \sqrt{16 - x^{2}}$ $int x^2sqrt(16-x^2)$ using trig substitutions?

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Answer 1 · 2016-09-05T23:24:04

$32 arcsin (\frac{x}{4}) - 8 sin (4 arcsin (\frac{x}{4})) + C$ $32arcsin(x/4)-8sin(4arcsin(x/4))+C$

Explanation:

Solving for:

$I = \int x^{2} \sqrt{16 - x^{2}} d x$ $I=intx^2sqrt(16-x^2)dx$

Make the substitution $x = 4 sin θ$ $x=4sintheta$ where $d x = 4 cos θ d θ$ $dx=4costhetad theta$ :

$I = \int 16 {sin}^{2} θ \sqrt{16 - 16 {sin}^{2} θ} 4 cos θ d θ$ $I=int16sin^2thetasqrt(16-16sin^2theta)4costhetad theta$

Bringing all the multiplicative constants out, including the $\sqrt{16} = 4$ $sqrt16=4$ :

$I = 256 \int {sin}^{2} θ cos θ \sqrt{1 - {sin}^{2} θ} d θ$ $I=256intsin^2thetacosthetasqrt(1-sin^2theta)d theta$

Since ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$ , we see that $cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta=sqrt(1-sin^2theta)$ :

$I = 256 \int {sin}^{2} θ {cos}^{2} θ d θ$ $I=256intsin^2thetacos^2thetad theta$

Here, since $sin 2 θ = 2 sin θ cos θ$ $sin2theta=2sinthetacostheta$ , we see that ${sin}^{2} 2 θ = 4 {sin}^{2} θ {cos}^{2} θ$ $sin^2 2theta=4sin^2thetacos^2theta$ :

$I = 64 \int 4 {sin}^{2} θ {cos}^{2} θ d θ$ $I=64int4sin^2thetacos^2thetad theta$

$I = 64 \int {sin}^{2} 2 θ d θ$ $I=64intsin^2 2theta d theta$

Note that, according to the cosine double angle formula, $cos 4 θ = 1 - 2 {sin}^{2} 2 θ$ $cos4theta=1-2sin^2 2theta$ , so ${sin}^{2} 2 θ = \frac{1 - cos 4 θ}{2}$ $sin^2 2theta=(1-cos4theta)/2$ :

$I = 32 \int (1 - cos 4 θ) d θ$ $I=32int(1-cos4theta)d theta$

$I = 32 \int d θ - 32 \int cos 4 θ d θ$ $I=32intd theta-32intcos4thetad theta$

Integrating both (substitution can be used for the second integral, or just see it for the reverse chain rule of $sin 4 θ$ $sin4theta$ that it is):

$I = 32 θ - 8 \int 4 cos 4 θ d θ$ $I=32theta-8int4cos4thetad theta$

$I = 32 θ - 8 sin 4 θ$ $I=32theta-8sin4theta$

Since $x = 4 sin θ$ $x=4sintheta$ , solving for $θ$ $theta$ gives $θ = arcsin (\frac{x}{4})$ $theta=arcsin(x/4)$ and $4 θ = 4 arcsin (\frac{x}{4})$ $4theta=4arcsin(x/4)$ . Thus:

$I = 32 arcsin (\frac{x}{4}) - 8 sin (4 arcsin (\frac{x}{4})) + C$ $I=32arcsin(x/4)-8sin(4arcsin(x/4))+C$

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How do you integrate $\int x^{2} \sqrt{16 - x^{2}}$ $int x^2sqrt(16-x^2)$ using trig substitutions?

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