$$\int{\sin^3x}\,\mathrm{d}x$$
I find that this integration is ambiguous since I could get the answer with different approaches. Are these answers are valid and true? Could someone tell me why and how? And also, is there any proof stating that these two method I use results the same value/answer?
Here how I work, please correct me if I'm wrong
First method :
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\sin x \cdot \sin^2x}\,\mathrm{d}x
\\ &= \int{\sin x (1 – \cos^2x)}\,\mathrm{d}x
\\& = \displaystyle\int{(\sin x – \sin x\cos^2x)}\,\mathrm{d}x
\\& = \dfrac{1}{3}\cos^3x – \cos x + C
\end{align}
Second method :
First, we know that $$\sin 3x = 3\sin x – 4\sin^3x$$
Therefore, $$\sin^3x = \dfrac{3}{4}\sin x – \dfrac{1}{4}\sin 3x$$
\begin{align}
\int{\sin^3x}\,\mathrm{d}x & = \int{\left(\frac{3}{4}\sin x – \frac{1}{4}\sin 3x\right)}\,\mathrm{d}x\\
& = \frac{1}{12}\cos 3x – \frac{3}{4}\cos x + C
\end{align}
Best Answer
$$\cos 3x =4\cos^3x -3\cos x$$ So, $$\frac{1}{12}\color{green}{\cos 3x} - \frac{3}{4}\cos x=\frac{1}{12}(\color{green}{4\cos^3x -3\cos x})-\frac{3}{4}\cos x$$ $$=\frac{1}{3}\cos^3x-\cos x$$
Hence both the answers are the same.