Getting different answers when computing $ \int \sin \theta \cos^3 \theta \, {\rm d}\theta $

Question

$$ \int \sin \theta \cos^3 \theta \, {\rm d}\theta $$

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As you can see, I get a correct answer using the first approach and an incorrect answer using the second approach

Answer 1

Your answers are equivalent and both correct.

It is possible to end up with two distinct-looking answers to an integral. Suppose that you find $F(x)$ and $G(x)$ are both indefinite integrals of $f(x)$. What this means is that:

• $F'=G'=f$
• Hence, by the mean value theorem, $F(x)=G(x)+k$ for some constant $k$

This is entirely fine. Remember, the goal of finding $\int f(x) \, \mathrm{d}x$ is simply to characterize all functions $F$ such that $F'=f$. From the second bullet point, if you have found one, you have found them all -- any antiderivative will differ from another by a constant.

You can see your answers are equivalent by the following:

• Graphing both of them, you see they're just a vertical shift apart:

• In fact, you should be able to set them equal to each other (with a constant $C$ on one side), and figure out the precise constant $C$ by which they differ.

• If you differentiate both of them, and manipulate trigonometric identities, you should be able to show that both of them have derivative $\sin(x) \cos^3(x)$.

Answer 2

Note that, for each $\theta\in\mathbb{R}$, \begin{align} -\frac{\cos^4\theta}4-\left(\frac{\sin^2\theta}2-\frac{\sin^4\theta}4\right)&=\frac14\left(\sin^4\theta-\cos^4\theta-2\sin^2\theta\right)\\ &=\frac14\left(\left(\sin^2\theta-\cos^2\theta\right)\left(\sin^2\theta+\cos^2\theta\right)-2\sin^2\theta\right)\\ &=\frac14\left(\sin^2\theta-\cos^2\theta-2\sin^2\theta\right)\\ &=\frac14\left(-\sin^2\theta-\cos^2\theta\right)\\ &=-\frac14, \end{align} and therefore your answers differ by a constant.