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  • Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 59 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 59 Maths textbook Solution.

Answers (1)

Answer: \pi

Hint: Use indefinite formula then put the limit to solve this integral

Given: \int_{1}^{2}\frac{1}{\sqrt{\left ( x-1 \right )\left ( 2-x \right )}}dx

Solution:

\int_{1}^{2} \frac{1}{\sqrt{(x-1)(2-x)}} d x=\int_{1}^{2} \frac{1}{\sqrt{2 x-x^{2}-2+x}} d x=\int_{1}^{2} \frac{1}{\sqrt{3 x-x^{2}-2}} d x\\

=\int_{1}^{2} \frac{1}{\sqrt{-\left(x^{2}-3 x+2\right)}} d x

=\int_{1}^{2} \frac{1}{\sqrt{-\left(x^{2}-2 x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2\right)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}-2 a b=(a-b)^{2}\right]

=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+2\right)}} d x

\begin{aligned} &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9-8}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}

Putting \left ( x-\frac{3}{2} \right )=t\Rightarrow dx=dt

When x=1 then t=1-\frac{3}{2}=\frac{2-3}{2}=\frac{-1}{2}

And When x=2  then t=2-\frac{3}{2}=\frac{4-3}{2}=\frac{1}{2}

Then \int_{1}^{2} \frac{1}{\sqrt{(x-1)(2-x)}} d x=\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}-t^{2}}} d t

=\left[\sin ^{-1}\left(\frac{t}{\frac{1}{2}}\right)\right]_{\frac{-1}{2}}^{\frac{1}{2}}                                                            \left[\int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1} \frac{x}{a}\right]

=\left[\sin ^{-1}(2 t)\right] \frac{-\frac{1}{2}}{2}

=\left[\sin ^{-1}\left(2 \times \frac{1}{2}\right)-\sin ^{-1}\left(2 \times \frac{-1}{2}\right)\right]                                                        [\sin (-\theta)=-\sin \theta]

=\sin ^{-1}(1)-\sin ^{-1}(-1)

=\sin ^{-1}(1)+\sin ^{-1}(1) \\

=2 \sin ^{-1}(1)                                                                    \left [ \sin ^{-1}=\frac{\pi }{2} \right ]

=2\times \frac{\pi }{2}

=\pi

 

Posted by

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