Question

Use Trig Substitution to find the integral of x^2/squareroot(9-x^2)dx

Answer 1

@satellite73 please help!

Answer 2

Let \(x=3\sin u\), so that \(dx=3\cos u~du\):
\[\int\frac{x^2}{\sqrt{9-x^2}}~dx~~~\Rightarrow~~~\int\frac{(3\sin u)^2}{\sqrt{9-(3\sin u)^2}}~(3\cos u~du)\]

Answer 3

\[\int\frac{(3\sin u)^2}{\sqrt{9-(3\sin u)^2}}~(3\cos u~du)=9\int\frac{\sin^2 u\cos u}{\sqrt{1-\sin ^2u}}~du\]

Answer 4

Keep on simplifying, you're left with
\[9\int\sin^2u~du\]
which can be rewritten using the following identity:
\[\sin^2u=\frac{1}{2}\left(1-\cos2u\right)\]

Answer 5

omg thank you!

Answer 6

how would i keep simplifying with the radical in the denominator?

Answer 7

\(1-\sin^2u=\cos^2u\) by the Pythagorean identity.

\(\sqrt{\cos^2u}=\cos u\).

The cosines disappear, leaving you with \(\sin^2u\).

Answer 8

so at the end would i multiply 9 by (1/2)(1-cos2u)?