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  • Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 31 Maths Textbook Solution.

Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 31 Maths Textbook Solution.

Answers (1)

Answer: x\sin ^{-1}\left ( \sqrt{x} \right )+\frac{1}{2}\left [ \sqrt{1-x}\sqrt{x}+\sin ^{-1} \sqrt{1-x}\right ]+c

Given:\int \sin ^{-1}\sqrt{xdx}

Hint: u\int vdx-\int \left ( \frac{d}{dx} u\int vdx\right )dx

Solution:

               \begin{aligned} &u=\int \sin ^{-1} \sqrt{x} d x \\ &=\int \sin ^{-1} \sqrt{x} \cdot 1 d x \\ &=\sin ^{-1} \sqrt{x} \int 1 d x-\int\left(\frac{d}{d x}\left(\sin ^{-1} \sqrt{x}\right) \int 1 d x\right) d x \end{aligned}

                 \begin{aligned} &{\left[\frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^{2}}}\right]} \\ &=x \sin ^{-1}(\sqrt{x})-\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \times \frac{1}{2 \sqrt{x}} x d x \\ &=x \sin ^{-1}(\sqrt{x})-\int \frac{\sqrt{x}}{2 \sqrt{1-(\sqrt{x})^{2}}} d x \end{aligned}

Now,

        \sqrt{1-\left ( \sqrt{x} \right )^{2}}=t

         \Rightarrow \sqrt{1-x}=t

         \Rightarrow 1-x=t^{2}

        \Rightarrow 1-t^{2}=x

        \Rightarrow \sqrt{x}=\sqrt{1-t^{2}}

Again

         \Rightarrow 1-x=t^{2}

         -dx=2tdt

         dx=-2tdt

   \left [ \int \sqrt{a^{2}-x^{2}}dx =\frac{1}{2} \left [ x\sqrt{a^{2}-x^{2}} a^{2}\sin ^{-1}\frac{x}{a}\right ]\right ]

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

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