Hello peeps,

What I need help on is clarifying if the answer for this integral is correct because I have did it in two ways and I'm not sure if both of them are correct.

$\displaystyle \int x^3\sqrt{9-x^2}\,dx$

$\displaystyle x = 3sin\Theta$

$\displaystyle dx = 9sin^2\Theta d\Theta\$

for the square root I have:

$\displaystyle \sqrt{9 - x^2}\ = \sqrt{9- 9 sin^2\Theta}\ = \sqrt{9(1-sin^2\Theta)}\ = \sqrt{9cos^2\Theta}\ = 3cos\Theta$

$\displaystyle

\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\

=

\int 729sin^3\Theta cos\Theta sin^2\Theta d\Theta\

$

From here I did the equation 2 ways, please let me know if one of the ways is incorrect.

1st way:

$\displaystyle

\int 27sin^3\Theta 3cos\Theta 9sin^2\Theta d\Theta\

$

$\displaystyle

u = sin\Theta

du = cos\Theta d\Theta

$

$\displaystyle

729\int u^3 u^2 du\

=

729\int u^5 du\

= 729 \frac{u^6}6 +C

= 729 \frac{sin^6}6 +C

= 729 \frac{\frac{x}{3}^6}{6} + C

$

2nd way

$\displaystyle

\int 729sin^3\Theta 3cos\Theta sin^2\Theta d\Theta\

=

\int 729sin^4\Theta cos\Theta sin\Theta d\Theta\

$

$\displaystyle

u = cos\Theta

du = -sin\Theta d\Theta

$

$\displaystyle

\int 729 (1-cos^2\Theta)^2 cos\Theta sin\Theta d\Theta\

=

-\int 729 u(1-u^2)^2 du

$

$\displaystyle

= -(1 - 2u^2 + u^4)u

= (1 + 2u^2 - u^4)u

= u + 2u^3 - u^5

$

$\displaystyle

= \frac{u^2}{2} + 2\frac{u^4}{4} - \frac{u^6}{6}

= \frac{cos\Theta^2}{2} + 2\frac{cos\Theta^4}{4} - \frac{cos\Theta^6}{6} + C

$

$\displaystyle

=729 \frac{\frac{\sqrt{9-x^2}^2}{3}}{2} + \frac{\frac{\sqrt{9-x^2}^4}{3}}{4} +

\frac{\frac{\sqrt{9-x^2}^6}{3}}{6} + C

$