# What am I missing when solving this integral with trigonometric substition?

The integral

$$\int\frac{x^2}{\sqrt{9-x^2}}dx$$

What I've tried

I've tried doing the substitution with

$$x^2=9\sin^2\theta$$ $$x=3\sin\theta$$ $$dx=3\cos\theta$$

my solving...

$$=\int\frac{27\sin^2\theta\cos\theta}{\sqrt{9-9\sin^2\theta}}d\theta$$ $$=27\int\frac{\sin^2\theta\cos\theta}{\sqrt{9(1-\sin^2\theta)}}d\theta$$ $$=9\int\frac{\sin^2\theta\cos\theta}{\sqrt{\cos^2\theta}}d\theta$$ $$=9\int\sin^2\theta d\theta$$ $$=\frac{9\theta}{2} - \frac{9\sin2\theta}{4}+ C$$

This is where I'm stuck.

The expected result $$\frac{(\sqrt{9-x^2})^3}{3}-9\sqrt{9-x^2}+C$$

Question

What errors have I done, or what am I missing ?

Please don't just throw the answer with fixed solving, I'd like to have some explanation so I don't reproduce errors.

Thanks.

I believe it should be $$=\int\frac{27\sin^{\color{red}2} \theta\cos\theta}{\sqrt{9-9\sin^2\theta}}d\theta$$ Then, notice that the integral becomes$$9\int \sin^{2} \theta\ d\theta$$ You've got it in terms of theta, so convert it back to x. Since $$x = 3\sin\theta$$ Theta is $$arcsin(\frac{x}{3}) =\theta$$ Substitute this and manipulate.

• Thanks, but I still don't get the correct result. Would you mind checking my edit ? Upvoted. – student Mar 12 '15 at 0:16
• @MathLearner Do that, and check here for ideas: en.wikipedia.org/wiki/Inverse_trigonometric_functions – Kugelblitz Mar 12 '15 at 0:21
• @MathLearner You can see that $\cos(\arcsin(x)) = \sqrt{1-x^2}$ , now you know that you've to expand the second term you've got with a trig identity, and substitute that formula I just mentioned earlier in this comment: sin(2x)=2sin(x)cos(x) – Kugelblitz Mar 12 '15 at 0:21
• how is cos(arcsin(x)) useful for me ? I end up with $$\frac{9(arcsin(\frac{x}{3}))}{2} - \frac{9\sin(\arcsin(\frac{x}{3}))}{4}$$ which simplify to $$\frac{9(arcsin(\frac{x}{3}))}{2} - \frac{4x}{3}$$ – student Mar 12 '15 at 0:29
• $9\sin(2\theta)=18\sin\theta\cos\theta$ – Kugelblitz Mar 12 '15 at 0:34

Notice that the integral is $$9\int \sin^{\color{red}2} \theta\ d\theta$$

• Thanks, but I still don't get the correct result. Would you mind checking my edit ? Upvoted. – student Mar 12 '15 at 0:15

$$x^2=9\sin^2\theta\neq 9\sin \theta$$ in your substitution.

• Thanks, but I still don't get the correct result. Would you mind checking my edit ? Upvoted. – student Mar 12 '15 at 0:15
• @MathLearner Note that you have to rewrite $\theta$ in terms of $x$. – Teoc Mar 12 '15 at 0:28