Question

Integrate the given equation ${\sin }^3(2x+1)$.

• A. $\frac{1}{2}$ $(-cos(2x+1\left)\right)$ + $\frac{1}{6}co{s}^3(2x+1)+c$
• B. $\frac{1}{3}$ $(-cos(2x+1\left)\right)$ + $\frac{1}{4}co{s}^3(2x+1)+c$
• C. $\frac{1}{2}$ $\left(cos\right(2x+1\left)\right)$ - $\frac{1}{3}co{s}^3(2x+1)+c$
• D. $(-cos(2x+1\left)\right)$ + ${\cos }^3(2x+1)+c$

Answer 1

The correct option is A $\frac{1}{2}$ $(-cos(2x+1\left)\right)$ + $\frac{1}{6}co{s}^3(2x+1)+c$

We are given, $I=\int {\sin }^3(2x+1)dx$

[Now, we know that $\sin 3\theta =3\sin \theta -4{\sin }^3\theta$]

So, ${\sin }^3\theta =\frac{1}{4}(3\sin \theta -\sin 3\theta )$

$I=\frac{1}{4}\left(3\sin \right(2x+1)-\sin (3(2x+1)\left)\right)dx$

$=\frac{1}{4}\int \left[3\sin \right(2x+1)-\sin (6x+3\left)\right]dx$

So, $I=\frac{1}{4}\left[\int 3\sin (2x+1)\right]dx-\frac{1}{4}\left[\int \sin (6x+3)\right]dx$

$=\frac{3}{4}\frac{(-\cos (2x+1\left)\right)}{2}-\frac{1}{4}\frac{(-\cos (6c+3\left)\right)}{6}+c$

$I=\frac{3}{8}(-\cos (2x+1)+\frac{1}{24}(\cos \left(3\right(2x+1\left)\right))+C$

Now, $\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

So, $I=\frac{3}{8}(-\cos (2x+1)+\frac{1}{24}(4{\cos }^3(2x+1))-3\cos (2x+1\left)\right)+c$

$=\frac{3}{8}(-\cos (2x+1\left)\right)+\frac{1}{6}{\cos }^3(2x+1)-\frac{1}{8}\cos (2x+1)+c$

$I=(\frac{3}{8}+\frac{1}{8})(-\cos (2x+1\left)\right)+\frac{1}{6}{\cos }^3(2x+1)+c$

$I=\frac{1}{2}(-\cos (2x+1\left)\right)+\frac{1}{6}{\cos }^3(2x+1)+c$.