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#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 56  Maths Textbook Solution.

Answer: $=2 \sqrt{x} \sin \sqrt{x}+\frac{4}{3} \cos \sqrt{x}-\frac{2 \sqrt{x} \sin ^{3} \sqrt{x}}{3}+\frac{2}{9}\left[\cos ^{3} \sqrt{x}\right]+c$

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

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

Solution:  $\sqrt{x}=t\Rightarrow x=t^{2}\Rightarrow dx=2tdt$

\begin{aligned} &\therefore \int \cos ^{3} \sqrt{x} d x=\int \cos ^{3} t .2 t d t \Rightarrow 2 \int t \cos ^{3} t d t \\ &=2 \int t \cos t \cos ^{2} t d t \Rightarrow 2 \int t \cdot \cos t\left(1-\sin ^{2} t\right) d t \\ &=2 \int\left(t \cos t-t \cos t \sin ^{2} t\right) d t \\ &=2\left[\int t \cos t d t-\int t \cos t \sin ^{2} t d t\right] \\ &=2\left[t \int \cos t d t-\int[1 \cdot \sin t d t]\right]-2\left[t \int \cos \sin ^{2} t d t-\int 1 \cdot\left(\frac{\sin ^{3} t}{3}\right) d t\right] \end{aligned}

Applying by parts

\begin{aligned} &=2\left(t \sin t-\int \sin t d t\right)-2\left(t \cdot \frac{\sin ^{3} t}{3}-\frac{1}{3} \int \sin ^{3} t d t\right) \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3} \int \sin t \cdot \sin ^{2} t d t \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3} \int \sin t \cdot\left(1-\cos ^{2} t\right) d t \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}\left[\int \sin t d t-\int \sin t \cos ^{2} t d t\right] \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}\left[-\cos t+\frac{\cos ^{3} t}{3}+c\right] \\ &=2 t \sin t+2 \cos t-\frac{2 t \sin ^{3} t}{3}+\frac{2}{3}[-\cos t]+\frac{2}{3}\left[\frac{\cos ^{3} t}{3}\right]+c \\ &=2 \sqrt{x} \sin \sqrt{x}+\frac{4}{3} \cos \sqrt{x}-\frac{2 \sqrt{x} \sin ^{3} \sqrt{x}}{3}+\frac{2}{9}\left[\cos ^{3} \sqrt{x}\right]+c \end{aligned}