im hoping i worked this out right; its long:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int x(81-x^2)^{5/2}dx[/tex]

the integral contains [itex]a^2-x^2[/itex], so i set [itex]x=asin\theta[/itex]. that would make [itex]x=9sin\theta[/itex] and [itex]dx=9cos\theta d\theta[/itex]:

[tex]\int 9sin\theta(81-81sin^2\theta)^{5/2}9cos\theta d\theta = \int 9sin\theta[81(1-sin^2\theta)]^{5/2}9cos\theta d\theta[/tex]

the integral now contains [itex]1-sin^2\theta =cos^2\theta[/itex]

[tex]\int 9sin\theta(9cos\theta)^59cos\theta d\theta[/tex]

i used u-sub by setting [itex]u=9cos\theta[/itex] and [itex]du=-9sin\theta d\theta[/itex].

[tex]-\int u^6 du=-\frac{u^7}{7}+C[/tex]

i plugged my u back in:

[tex]-\frac{(9cos\theta)^7}{7}+C[/tex]

then, i drew my little triangle.

[tex]cos\theta=\frac{\sqrt{81-x^2}}{9}[/tex]

i plugged that into the [itex]cos\theta[/itex] and simplified and came out with:

[tex]-\frac{(\sqrt{81-x^2})^7}{7}+C[/tex]

that's my answer

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# Tricky trig substitution

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