Equivalent integral results for $\int \sin^3(x) \rm dx$

Question

I am currently revising for the math subject GRE and when I check my work against Wolfram Alpha and https://www.integral-calculator.com, my solution is (almost) always of the latter form rather than the former.

So for example, the latter gives

$$\int \sin^3(x) dx = \cos(x) -\frac{1}{3}\cos^3(x)+C$$

while Wolfram Alpha gives

$$\int \sin^3(x) dx = -\frac{3}{4}\cos(x) +\frac{1}{4}\cos(3x)+C$$

while these are equivalent, I think $most$ people would naturally generate the first solution..... therefore, I assume the solution to appear on the GRE would be of the first form rather than the second.

(I guess Wolfram alpha just prefers this form since we can take derivatives and integrals of the output easier, for example).

Regardless.... in preparing for the exam, should I comfortably switch between these forms or suppose that I get the first form but see nothing like it in the options.... how can I quickly compare my solution to see that it is correct?

Answer 1

Math Lover is right in that if you are ever faced with a multiple choice problem of the form "find the antiderivative", it is almost always best to take the derivative of each answer and compare. But still, I'd say you should at least know the angle addition formulas, namely $$\sin(\alpha+\beta) = \sin(\alpha)\cos(\beta)+\sin(\beta)\cos(\alpha)$$ $$\cos(\alpha+\beta) = \cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$ Using these, one can quickly derive an expansion of $\cos(3x)$ and thus comfortably switch between the two forms you have stated.

Answer 2

To the OP: For your first integral, the solution is: $$\int{\sin^3(x)}dx=-\cos(x)+\frac{1}{3}\cos^3(x)+C$$ Notice that the signs in your solution are reversed.

Secondly, by using the identity: $${\cos(3x)}\equiv {4\cos^3(x)-3\cos(x)}$$ And substituting for $$\cos^3(x) \equiv \frac{1}{4}{\left[\cos(3x)+3\cos(x)\right]}$$ You will get, the solution from Wolfram Alpha which is: $$\int{\sin^3(x)}dx=-\cos(x)+\frac{1}{3}{\left(\frac{1}{4}{\left[\cos(3x)+3\cos(x)\right]}\right)}=-\frac{3}{4}\cos(x)+\frac{1}{12}\cos(3x)+C$$

Both solutions are the same, but the second is cleaner, since there are no powers of $\cos(x)$.

Also, notice in your second integral that you have $\frac{1}{4}\cos(3x)$ when it should be $\frac{1}{12}\cos(3x)$