$\int x^{2} \sqrt{9 - x^{2}} d x$ $\intx^2\sqrt(9-x^2)dx$ ?

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$\int x^{2} \sqrt{9 - x^{2}} d x$ $\intx^2\sqrt(9-x^2)dx$ ?

Symbolab does it a certain way, but is there a different method of approach?

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Answer 1 · 2018-04-30T05:25:42

$\int x^{2} \sqrt{9 - x^{2}} d x = \frac{81}{8} [arcsin (\frac{x}{3}) - \frac{1}{4} sin (4 arcsin (\frac{x}{3}))]$ $intx^2sqrt(9-x^2)dx=81/8[arcsin(x/3)-1/4sin(4arcsin(x/3))]$

Explanation:

I also follow similar way but my steps may be slightly more explained and are as follows.

Let $x = 3 sin u$ $x=3sinu$ then $d x = 3 cos u d u$ $dx=3cosudu$ and this means $u = arcsin (\frac{x}{3})$ $u=arcsin(x/3)$

and $\int x^{2} \sqrt{9 - x^{2}} d x = \int 9 {sin}^{2} u \sqrt{9 - 9 {sin}^{2} u} \cdot 3 cos u d u$ $intx^2sqrt(9-x^2)dx=int9sin^2usqrt(9-9sin^2u)*3cosudu$

= $81 \int {sin}^{2} u {cos}^{2} u d u$ $81intsin^2ucos^2udu$

= $\frac{81}{4} \int {sin}^{2} 2 u d u$ $81/4intsin^2 2udu$

= $\frac{81}{4} \int \frac{1 - cos 4 u}{2} d u$ $81/4int(1-cos4u)/2du$

= $\frac{81}{8} [\int d u - \int cos 4 u d u]$ $81/8[intdu-intcos4udu]$

= $\frac{81}{8} [u - \frac{sin 4 u}{4}]$ $81/8[u-(sin4u)/4]$

= $\frac{81}{8} [arcsin (\frac{x}{3}) - \frac{1}{4} sin (4 arcsin (\frac{x}{3}))]$ $81/8[arcsin(x/3)-1/4sin(4arcsin(x/3))]$

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$\int x^{2} \sqrt{9 - x^{2}} d x$ $\intx^2\sqrt(9-x^2)dx$ ?

Symbolab does it a certain way, but is there a different method of approach?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

∫x2√9−x2dx\intx^2\sqrt(9-x^2)dx?

Symbolab does it a certain way, but is there a different method of approach?

1 Answer

Explanation:

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$\int x^{2} \sqrt{9 - x^{2}} d x$ $\intx^2\sqrt(9-x^2)dx$ ?