Question

Evaluate :$\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}}dx$

• A. $x+2\sin x+\sin 2x+c$
• B. $x+2\cos x+\sin 2x+c$
• C. $x-2\sin x+\sin 2x+c$
• D. $x+2\sin x-\sin 2x+c$

Answer 1

The correct option is A

$x+2\sin x+\sin 2x+c$

Explanation for the correct option:

Given: $\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}}dx$

Let $I=\int \frac{\sin \frac{5x}{2}}{\sin \frac{x}{2}}dx$

$$\begin{gathered} =\int \frac{2\sin \left(\frac{5x}{2}\right)\cos \left({\displaystyle \frac{x}{2}}\right)}{\sin x}dx \\ =\int \frac{\sin 3x+\cos 2x}{\sin x}dx\left[\because \sin a+\sin b=\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)\right] \\ =\int \frac{3\sin x-4{\sin }^{3}x-2\sin x\cos x}{\sin x}dx\left[\because \sin 3x=3\sin x-4{\sin }^{3}x\right] \\ =\int \frac{3\sin x}{\sin x}-\frac{4{\sin }^{3}x}{\sin x}-\frac{2\sin x\cos x}{\sin x}dx \\ =\int \left(3-4{\sin }^{2}x-2\cos x\right)dx \\ =\int \left(3-2\left(1-\cos 2x\right)-2\cos x\right)dx\left[\because 1-\cos 2x=\frac{{\sin }^{2}x}{2}\right] \\ =\int \left(3-2\cos 2x-2\cos x\right)dx \\ =x+\frac{2\sin 2x}{2}+\frac{2\sin x}{2}+c \\ =x+\sin 2x+\sin x+c \end{gathered}$$

Therefore the value of integration is $x+2\sin x+\sin 2x+c$

Hence, the correct option is (A)