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

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

Answers (1)

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

Posted by

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