Find the indefinite integral via substitution rule

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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

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

Answer 4

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