problem doing this trig integral $\int \frac{x^3dx}{\sqrt{9-x^2}}$

Question

$$\int \frac{x^3dx}{\sqrt{9-x^2}}$$ $$x = 3\sin\theta$$ $$dx =3\cos\theta d\theta$$ $$\int \frac{(3\sin\theta)^33\cos\theta d\theta}{3\cos\theta }$$ $$\int 27\sin^3\theta d\theta$$ $$-27\cos\theta + 27\frac{9\cos^3}{3}$$

Replacing cosine by $\frac{\sqrt{9-x}}{3}$ does not give the correct answer what did I get wrong?

correct answer: $$-\frac{\sqrt{9-x^2}(x^2+18)}{3}$$

I don't know about whether there's any step that is wrong, but is there a reason you used a trig substitution rather than just $u = 9-x^2$? — Brian Moehring, Jul 06 '23 at 01:16
All right, well, your first error, then, is that $3^3 = 27 \neq 9$. — Brian Moehring, Jul 06 '23 at 01:18
You may want to give the answer you reached then. If you really replaced $\cos\theta$ with $\frac{\sqrt{9-x}}{3}$, then that's another error, but I can't tell if it affected your answer or whether it's a typo. — Brian Moehring, Jul 06 '23 at 01:21
Also: it may look different but still be correct. If you got an answer of $(x+4)^2$ and I said the answer is $x^2+8x+16$, we're not wrong. — Randall, Jul 06 '23 at 01:23
https://math.stackexchange.com/questions/960298/integrate-int-fracx3dx-sqrt1x2 and https://math.stackexchange.com/questions/2148572/solving-an-indefinite-integral — lab bhattacharjee, Jul 06 '23 at 05:07
Typesetting tip: Typing $\sin\theta$ , $\cos\theta$ , $\tan\theta$ , $\csc\theta$ , $\sec\theta$ , and $\cot\theta$ produces $\sin\theta$, $\cos\theta$, $\tan\theta$, $\csc\theta$, $\sec\theta$, and $\cot\theta$, respectively. — N. F. Taussig, Jul 06 '23 at 11:47

Brian Moehring · Accepted Answer · 2023-07-06T01:53:53.837

First and foremost, this is not an integral for which trigonometric substitution is the best way to resolve it. We could instead just use $u = 9 - x^2$ to instead find $x^2 = 9-u$ and $x\,dx = -\frac12\,du$.

This gives $$\int\frac{x^3}{\sqrt{9-x^2}}\,dx = \int\frac{x^2}{\sqrt{9-x^2}}\cdot x\,dx = \int\frac{9-u}{\sqrt{u}}\left(-\frac12\right) \,du = -\frac12 \int\left(9u^{-1/2} - u^{1/2}\right)\,du$$

and you should be able to finish.

That said, we certainly can use a trig substitution as you have done.

It gives $$\int\frac{x^3}{\sqrt{9-x^2}}\,dx = \int 27\sin^3\theta\,d\theta = 27\int(1-\cos^2\theta)\sin\theta\,d\theta$$

Here, we use the substitution $u = \cos\theta$ to write (I know I've already harped on it, but the fact the standard method of evaluation uses substitution again to get rid of all the trig functions is a good indication we should've used a different substitution) $$-27\int (1-u^2)\,du = -27\cos\theta + 9\cos^3\theta$$

Substituting $\cos\theta = \frac{\sqrt{9-x^2}}{3}$: $$-27\left(\frac{\sqrt{9-x^2}}{3}\right) + 9\left(\frac{\sqrt{9-x^2}}{3}\right)^3 = -\frac13\sqrt{9-x^2}\cdot \left(27 - (9-x^2)\right) = -\frac13\sqrt{9-x^2}(18+x^2)$$

Which errors does this show in your attempt?

Note that when integrating $27\sin^3\theta$ the coefficient of $\cos^3\theta$ is $9$.
You have at times used $\cos\theta = \frac{\sqrt{9-x}}{3}$ (the wrong one) and at times $\cos\theta = \frac{\sqrt{9-x^2}}{3}$ (the correct one). It's not clear why you have this inconsistency.
You still seem to have a point you've written $3^3 = 9$.

Lai · Answer 2 · 2023-07-06T02:16:41.507

Correction of OP: $$\cos \theta =\sqrt{1-\sin ^2 \theta} =\sqrt{1-\frac{x^2}{9}} =\frac{1}{3} \sqrt{9-x^2}$$

$$ \begin{aligned} I & =27 \int \sin ^3 \theta d \theta \\ & =-27 \int\left(1-\cos ^2 \theta\right) d(\cos \theta) \\ & =-27\left(\cos \theta-\frac{\cos ^3 \theta}{3}\right) \\ & =-9 \cos \theta\left(3-\cos ^2 \theta\right)\\&=-\frac{\sqrt{9-x^2}}{3}\left(x^2+18\right)+C \end{aligned} $$

Suggested solution $$ \begin{aligned}\int \frac{x^3}{\sqrt{9-x^2}} d x = & -\int x^2 d \sqrt{9-x^2} \\ = & -x^2 \sqrt{9-x^2}-\int \sqrt{9-x^2} d\left(9-x^2\right) \\ = & -x^2 \sqrt{9-x^2}-\frac{2\left(9-x^2\right)^{\frac{3}{2}}}{3}+C \\ = & -\frac{\sqrt{9-x^2}}{3}\left[3 x^2+2\left(9-x^2\right)\right]+C \\ = & -\frac{\sqrt{9-x^2}}{3}\left(x^2+18\right)+C \end{aligned} $$

problem doing this trig integral $\int \frac{x^3dx}{\sqrt{9-x^2}}$

2 Answers2