0
$\begingroup$

Find the indefinite integral $\int_{} (\cos^3x)(\sin x)\mathrm dx$

Here is my work.

1) Pick the $u, v$ values:

$$u = \cos x, \mathrm du = -\sin x$$ $$v = x, \mathrm dv = 1$$

2) Substitute $u, v$ values into integral

$$= \int_{} (u)^3(\sin v)(-\sin x)\tag1$$

3) Integrate (Find the antiderivative)

$$= \frac{u^4}{4}(\cos v)(-\cos x) $$

4) Put substitutes into the antiderivative

$$\frac{1}{4} (\cos x)^4 (\cos x)(-\cos x) = \textbf{$-\frac{(\cos x)^6}{4}$} + C = {-\frac{\cos^6x}{4}} + C $$

However the textbook says the answer is $-\frac{\cos^4x}{4} + C$

I am confused, where did I go wrong? I felt like I followed the substitution rule correctly. I am not sure if I used the substitution rule on $\sin x$ correctly in this context.

$\endgroup$
1
  • $\begingroup$ It looks like you've mixed up substitution with integration by parts. $\endgroup$ – J.G. Feb 11 '19 at 16:11
4
$\begingroup$

Substitute $u = \cos x$ and hence $du = -\sin x \: \mathrm{d}x$. Your integral becomes $-\int u^3 \; \mathrm{d}u = -\displaystyle\frac{u^4}{4} = -\frac{\cos^4 x}{4} + C$ which is the desired answer.

$\endgroup$
3
  • $\begingroup$ why is it okay to ignore the $sinx$ in the original indefinite integral? $\endgroup$ – Evan Kim Feb 11 '19 at 16:50
  • 1
    $\begingroup$ We are including $\sin x$ in $\mathrm{d}u = \sin x \mathrm{d}x$. Essentially, we are not ignoring. $\endgroup$ – Abhay Hegde Feb 11 '19 at 17:09
  • $\begingroup$ Please note that it should be $\mathrm{d}u = - \sin x \: \mathrm{d}x$. Sorry for the typo. $\endgroup$ – Abhay Hegde Feb 11 '19 at 17:24
2
$\begingroup$

Your second step is wrong. When you substitute $u=\cos x$, you should get $\mathrm du=-\sin x\; \mathrm dx$. So your integral becomes

$$\int \cos^3x\sin x \;\mathrm dx=-\int u^3 \; \mathrm du.$$

$\endgroup$
3
  • $\begingroup$ It feels like I am just ignoring $sinx$ that was in the original indefinite integral. Why am I allowed to do that when doing substitution? $\endgroup$ – Evan Kim Feb 11 '19 at 16:51
  • $\begingroup$ @EvanKim No, we are not ignoring $\sin x $. $\sin x $ is included in $\mathrm du $. (Note that $-\mathrm du=\sin x\; \mathrm dx $.) $\endgroup$ – Shivering Soldier Feb 11 '19 at 16:53
  • 1
    $\begingroup$ Oh, so essentially it is already accounted for? That makes more sense I guess. If $sinx$ was not there, it would not be included and therefore we would not be able to evaluate the indefinite integral $\endgroup$ – Evan Kim Feb 11 '19 at 16:56
1
$\begingroup$

Substitute $$t=\cos(x)$$ then we get $$dt=-\sin(x)dx$$

$\endgroup$
0
$\begingroup$

$$I=\int\cos^3(x)\sin(x)dx$$ $$u=\cos(x),\,dx=\frac{du}{-\sin(x)}$$ $$I=\int\cos^3(x)\sin(x)\frac{du}{-\sin(x)}=\int-u^3du$$

$\endgroup$

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.