Question

Let $f\left(x\right)={\sin }^{3}x+{\sin }^3(x+\frac{2\pi }{3})+{\sin }^3(x+\frac{4\pi }{3})$ then the primitive of $f\left(x\right)$ w.r.t $x$ is where $C$ is an arbitrary constant

• A. $-\frac{3\sin 3x}{4}+C$
• B. $-\frac{3\cos 3x}{4}+C$
• C. $\frac{\sin 3x}{4}+C$
• D. $\frac{\cos 3x}{4}+C$

Answer 1

Answer: (D)

$\frac{\cos 3x}{4}+C$

Let
$I=\int ({\sin }^3\left(x\right)+{\sin }^3(x+\left(\frac{2\pi }{3}\right))+{\sin }^3(x+\left(\frac{4\pi }{3}\right)))dx=\int ({\sin }^3\left(x\right)-{\cos }^3(\frac{\pi }{6}-x)+{\cos }^3(\frac{\pi }{6}+x))dx=\int {\sin }^3\left(x\right)dx+\int {\cos }^3(\frac{\pi }{6}+x)dx-\int {\cos }^3(\frac{\pi }{6}-x)dx$
U\sin g reduction formula
$\int {\sin }^m\left(x\right)dx=-\frac{\cos \left(x\right){\sin }^{m-1}\left(x\right)}{m}+\frac{m-1}{m}\int {\sin }^{-2+m}\left(x\right)dx$
where m=3, we get
$I=-\frac{1}{3}{\sin }^2\left(x\right)\cos \left(x\right)+\frac{2}{3}\int \sin \left(x\right)dx+\int {\cos }^3(\frac{\pi }{6}+x)dx-\int {\cos }^3(\frac{\pi }{6}-x)dx=-\frac{2\cos \left(x\right)}{3}-\frac{1}{3}{\sin }^2\left(x\right)\cos \left(x\right)+\int {\cos }^3(\frac{\pi }{6}+x)dx-\int {\cos }^3(\frac{\pi }{6}-x)dx$
For $\int {\cos }^3(\frac{\pi }{6}+x)dx$
Substitute $\frac{\pi }{6}+x=t$ and $dx=dt$
And u\sin g reduction formula
$\int {\cos }^m\left(x\right)dx=\frac{\sin \left(x\right){\cos }^{m-1}\left(x\right)}{m}+\frac{m-1}{m}\int {\cos }^{-2+m}\left(x\right)dx$
We get,
$\int {\cos }^3(\frac{\pi }{6}+x)dx=\frac{2\sin (\frac{\pi }{6}+x)}{3}+\frac{1}{3}\sin (\frac{\pi }{6}+x){\cos }^2(\frac{\pi }{6}+x)$
Similarly
$\int {\cos }^3(\frac{\pi }{6}-x)dx=\frac{2\sin (\frac{\pi }{6}-x)}{3}+\frac{1}{3}\sin (\frac{\pi }{6}-x){\cos }^2(\frac{\pi }{6}-x)$
Substituting these values in I and simplifying, we get
$I=\frac{1}{4}\cos \left(3x\right)+C$