Find the indefinite integral via substitution rule

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.

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

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.

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

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

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

• 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? – Evan Kim Feb 11 '19 at 16:51
• @EvanKim No, we are not ignoring $\sin x$. $\sin x$ is included in $\mathrm du$. (Note that $-\mathrm du=\sin x\; \mathrm dx$.) – Shivering Soldier Feb 11 '19 at 16:53
• 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 – Evan Kim Feb 11 '19 at 16:56

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

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