Question

Find the integral of the given function w.r.t - x

$y=\frac{si{n}^2\sqrt{x}}{2\sqrt{x}}$

• A. $\frac{1}{2}{x}^3+\frac{1}{3}sin3x+c$
• B. $\frac{1}{2}\sqrt{x}-\frac{1}{4}sin2\sqrt{x}+c$
• C. $\frac{1}{2}\sqrt{x}-\frac{1}{4}sin\sqrt{x}+c$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{2}\sqrt{x}-\frac{1}{4}sin2\sqrt{x}+c$

$I=\int ydx=\int \frac{si{n}^2\sqrt{x}}{2\sqrt{x}}dx$

Let $u=\sqrt{x},\frac{du}{dx}=\frac{1}{2\sqrt{x}}$

$\Rightarrow du=\frac{dx}{2\sqrt{x}}$ ......(i)

$\therefore I=\int si{n}^{2}udu$(substituting (i))

= $\int \frac{1-cos2u}{2}du(\because cos2a=1-2si{n}^{2}a)$

= $\int \frac{1}{2}du-\frac{1}{2}\int cos2udu$

Let $t=2u\Rightarrow \frac{dt}{du}=2\Rightarrow du=\frac{dt}{2}$

$\therefore I=\int \frac{1}{2}du-\frac{1}{2}\int \frac{costdt}{2}$

= $\frac{1}{2}u+c_1-\frac{1}{4}\int costdt$

= $\frac{1}{2}u+c_1-\frac{1}{4}sint+c_2$

= $\frac{1}{2}u-\frac{1}{4}sint+c_3$

= $\frac{1}{2}\sqrt{x}-\frac{1}{4}sin2u+c_3$

= $\frac{1}{2}\sqrt{x}-\frac{1}{4}sin2\sqrt{x}+c_3$