#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 54 Maths Textbook Solution.

Answer: $-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c$

Hint: Put $\sqrt{x}=t$

Given: $\int \sin ^{3}\sqrt{xdx}$

Solution: Put

$\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}$

$\Rightarrow dx=2tdt$

\begin{aligned} &\therefore \int \sin ^{3} \sqrt{x} d x=2 \int t \sin ^{3} t d t \\ &{\left[\begin{array}{l} \therefore \sin 3 t=3 \sin t-4 \sin ^{3} t \\ \sin ^{3} t=\frac{3 \sin t-\sin 3 t}{4} \end{array}\right]} \\ &\Rightarrow 2 \int \frac{t[3 \sin t-\sin 3 t]}{4} d t \\ &\Rightarrow \frac{3}{2} \int t \sin t d t-\frac{1}{2} \int t \sin 3 t d t \end{aligned}

Applying by parts,

\begin{aligned} &=\frac{3}{2}\left[t(-\cos t)-\int(-\cos t) d t\right]-\frac{1}{2}\left[t\left(\frac{-\cos 3 t}{3}\right)-\int\left(\frac{-\cos 3 t}{3}\right) d t\right] \\ &=\frac{3}{2}[-t \cos t+\sin t]-\frac{1}{2}\left[\frac{-t \cos 3 t}{3}+\frac{1}{3} \frac{\sin 3 t}{3}\right]+c \\ &=-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c \end{aligned}

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Answer: $-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c$

Hint: Put $\sqrt{x}=t$

Given: $\int \sin ^{3}\sqrt{xdx}$

Solution: Put

$\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}=\frac{dt}{dx}$

$\Rightarrow dx=2tdt$

\begin{aligned} &\therefore \int \sin ^{3} \sqrt{x} d x=2 \int t \sin ^{3} t d t \\ &{\left[\begin{array}{l} \therefore \sin 3 t=3 \sin t-4 \sin ^{3} t \\ \sin ^{3} t=\frac{3 \sin t-\sin 3 t}{4} \end{array}\right]} \\ &\Rightarrow 2 \int \frac{t[3 \sin t-\sin 3 t]}{4} d t \\ &\Rightarrow \frac{3}{2} \int t \sin t d t-\frac{1}{2} \int t \sin 3 t d t \end{aligned}

Applying by parts,

\begin{aligned} &=\frac{3}{2}\left[t(-\cos t)-\int(-\cos t) d t\right]-\frac{1}{2}\left[t\left(\frac{-\cos 3 t}{3}\right)-\int\left(\frac{-\cos 3 t}{3}\right) d t\right] \\ &=\frac{3}{2}[-t \cos t+\sin t]-\frac{1}{2}\left[\frac{-t \cos 3 t}{3}+\frac{1}{3} \frac{\sin 3 t}{3}\right]+c \\ &=-\frac{3}{2} \sqrt{x} \cos \sqrt{x}+\frac{3}{2} \sin \sqrt{x}+\frac{1}{6}\left[\sqrt{x} \cos 3 \sqrt{x}-\frac{\sin 3 \sqrt{x}}{3}\right]+c \end{aligned}