- #1
p53ud0 dr34m5
- 94
- 0
im hoping i worked this out right; its long:
[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
[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