<s>[INST] The result of $\int{\sin^3x}\,\mathrm{d}x$ $$\int{\sin^3x}\,\mathrm{d}x$$
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}
 [/INST] Since $$\cos^3(x)=\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)$$
Your first integral becomes $$\int \sin^3(x)dx=\dfrac{1}{3}\cos^3x - \cos x + C$$
$$=\frac{1}{3}\big[\frac{3}{4}\cos(x)+\frac{1}{4}\cos(3x)\big]-\cos(x)+C$$
$$=\frac{1}{12}\cos(3x)-\frac{3}{4}\cos(x)+C$$

Note that the constant of integration are not necessarily the same. For example using $u$-substitutions for the denominator we have $$\int \frac{4x}{4x^2+7}dx=\frac{1}{2}\ln(4x^2+7)+C_{1}$$
$$\int \frac{x}{x^2+\frac{7}{4}}dx=\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}$$
Here we have $C_{2}=C_{1}+\frac{1}{2}\ln(4)$ since they are constants. Indeed we have $$\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}=\frac{1}{2}\ln(x^2+\frac{7}{4})+\frac{1}{2}\ln(4)+C_{1}$$
$$=\frac{1}{2}\big[\ln(x^2+\frac{7}{4})+\ln(4)\big]+C_{1}$$
$$=\frac{1}{2}\ln(4(x^2+\frac{7}{4}))+C_{1}$$
$$=\frac{1}{2}\ln(4x^2+7)+C_{1}.$$
</s>