1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks, but I still don't get the correct result. Would you mind checking my edit ? Upvoted. $\endgroup$ – student Mar 12 '15 at 0:16
  • $\begingroup$ @MathLearner Do that, and check here for ideas: en.wikipedia.org/wiki/Inverse_trigonometric_functions $\endgroup$ – Kugelblitz Mar 12 '15 at 0:21
  • $\begingroup$ @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) $\endgroup$ – Kugelblitz Mar 12 '15 at 0:21
  • $\begingroup$ 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 $$\frac{9(arcsin(\frac{x}{3}))}{2} - \frac{4x}{3}$$ $\endgroup$ – student Mar 12 '15 at 0:29
  • $\begingroup$ $9\sin(2\theta)=18\sin\theta\cos\theta$ $\endgroup$ – Kugelblitz Mar 12 '15 at 0:34
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Thanks, but I still don't get the correct result. Would you mind checking my edit ? Upvoted. $\endgroup$ – student Mar 12 '15 at 0:15
1
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.