#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 30 Maths Textbook Solution.

Answer: $\frac{-\cos ^{4\left ( \cos ^{-1}\sqrt{x} \right )}}{2}+c$

Hint: Let

$x=\cos ^{2}t$

Given:

$\int \sec ^{-1}\sqrt{xdx}$                                                                    ................(1)

Solution:

Assume that $\sqrt{x}=t$                                                                        ...............(ii)

$\Rightarrow x=t^{2}$

Differentiating w.r.t to ‘t’

$dx=2tdt$                                                            .................(iii)

Now substituting values from equation (ii) & equation (iii)

\begin{aligned} I &=\int \sec ^{-1} t(2 t d t) \\ I &=2 \int \sec ^{-1} t . t d t \\ &=2\left\{\sec ^{-1} t \int t d t-\int\left(\frac{d\left(\sec ^{-1} t\right)}{d t}\left(\int t d t\right)\right) d x\right\} \end{aligned}

Here, use integration Identity

$\left [ \int x^{n}dx=\frac{x^{n+1}}{n+1}+c \right ]$

$\begin{gathered} I=2\left\{\sec ^{-1} t \frac{t^{2}}{2}-\int\left(\frac{1}{t \sqrt{t^{2}-1}}\left(\frac{t^{2}}{2}\right)\right) d t\right\}+c \\ I=t^{2} \sec ^{-1} t-\int \frac{t}{\sqrt{t^{2}-1}} d t+c \end{gathered}$                        ...........(iv)

Assume

$t^{2}-1=y$                                                                                                                   ...........(v)

Differentiate w.r.t ‘y’

$2tdt=dy$

$tdt=\frac{dy}{2}$                                                                                                                           ............(vi)

Substituting values from equation (v) and equation (vi) in integral of equation (iv)

\begin{aligned} &I=t^{2} \sec ^{-1} t-\int \frac{t}{\sqrt{y}} \frac{d y}{2}+c \\ &I=t^{2} \sec ^{-1} t-\frac{1}{2} \int y^{-\frac{1}{2}} d y+c \end{aligned}

Use integration identity

$\left [ \int x^{n}dx=\frac{x^{n+1}}{n+1} +c\right ]$

\begin{aligned} &I=t^{2} \sec ^{-1} t-\frac{1}{2} \frac{y^{\frac{1}{2}}}{\frac{1}{2}}+c \\ &I=t^{2} \sec ^{-1} t-\sqrt{y}+c \end{aligned}

Substitute the value of y from equation (v)

$I=t^{2}\sec ^{-1}t-\sqrt{t^{2}-1}+c$

Substitute the value of ‘t’ from equation (ii)

$I=x\sec ^{-1}\sqrt{x}-\sqrt{x-1}+c$

So,

$\int \sec ^{-1} \sqrt{x} d x=x \sec ^{-1} \sqrt{x}-\sqrt{x-1}+c$