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}

  • 2
    $\begingroup$ Prove that these answers are the same, by proving that $\frac{1}{12} \cos(3x) - \frac 34 \cos(x) -(\frac 13 \cos^3 x - \cos x)$ is a constant. $\endgroup$ Sep 30, 2020 at 9:47
  • $\begingroup$ Should it equal to zero? How to do that? Could you give me some details, please? $\endgroup$
    – Wilory Lu
    Sep 30, 2020 at 9:49
  • $\begingroup$ Yes, it should equal $0$. Substitute $x = \frac \pi 2$, then all terms are zero. Use the triple angle formula. $\endgroup$ Sep 30, 2020 at 9:50
  • $\begingroup$ Wow, I also see that when $x = 0$, the result holds. Thanks. $\endgroup$
    – Wilory Lu
    Sep 30, 2020 at 9:52
  • $\begingroup$ You are welcome! $\endgroup$ Sep 30, 2020 at 9:52

3 Answers 3


$$\cos 3x =4\cos^3x -3\cos x$$ 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.


Yes, they are both valid and true. Actually,$$(\forall x\in\Bbb R):\frac13\cos^3(x)-\cos(x)=\frac1{12}\cos(3x)-\frac34\cos(x)$$since$$(\forall x\in\Bbb R):\cos(3x)=4\cos^3(x)-3\cos(x).$$


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 $$\frac{1}{2}\ln(x^2+\frac{7}{4})+C_{2}=\frac{1}{2}\ln(x^2+\frac{7}{4})+\frac{1}{2}\ln(4)+C_{1}$$ $$=\frac{1}{2}\big[\ln(x^2+\frac{7}{4})+\ln(4)\big]+C_{1}$$ $$=\frac{1}{2}\ln(4(x^2+\frac{7}{4}))+C_{1}$$ $$=\frac{1}{2}\ln(4x^2+7)+C_{1}.$$


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