# Trigonometric substitution

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• November 27th 2008, 08:20 PM
Link88
Trigonometric substitution
Can someone help me with this trig subst please x^3/Sqrt(9-x^2) with respect to dx. Thanks
• November 27th 2008, 09:12 PM
Soroban
Hello, Link88!

Quote:

Integrate: . $\int\frac{x^3}{\sqrt{9-x^2}}\,dx$
Let: $x \:=\:3\sin\theta \quad\Rightarrow\quad dx \:=\:3\cos\theta\,d\theta$ . . . and: . $\sqrt{9-x^2} \:=\;3\cos\theta$

Substitute: . $\int\frac{27\sin^3\!\theta}{3\cos\theta}\,3\cos\th eta\,d\theta \;=\;27\int\sin^3\!\theta\,d\theta \;=\;27\int(\sin^2\!\theta)(\sin\theta\,d\theta)$

. . $= \;27\int(1 - \cos^2\!\theta)(\sin\theta\,d\theta) \;=\;27\bigg[\int\sin\theta\,d\theta - \int\cos^2\!\theta(\sin\theta\,d\theta) \bigg]$

. . $= \;27\bigg[-\cos\theta + \frac{1}{3}\cos^3\!\theta\bigg] + C \;=\;9\cos\theta\left(\cos^2\!\theta - 3\right) + C$

Back-substitute: . $\sin\theta \,=\,\frac{x}{3}\quad\Rightarrow\quad \cos\theta \,=\,\frac{\sqrt{9-x^2}}{3}$

and we have: . $9\cdot\frac{\sqrt{9-x^2}}{3}\left(\frac{9-x^2}{9} - 3\right) + C \;=\;3\sqrt{9-x^2}\left(\frac{9-x^2-27}{9}\right) + C$

. . $= \;\;\sqrt{9-x^2}\left(\frac{-x^2-18}{3}\right) \;\;=\;\;\boxed{-\frac{1}{3}\sqrt{9-x^2}\,(x^2+18) + C}$