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  • Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 43 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 43 Maths Textbook Solution.

Answers (1)

Answer:

\pi

Hint: Use indefinite formula and put the limits to solve this integral

Given:

\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx

Solution:

\int_{0}^{4}\frac{1}{\sqrt{4x-x^{2}}}dx=\int_{0}^{4}\frac{1}{\sqrt-{\left ( x^{2}-2.2x+\left ( 2 \right )^{2}-\left ( 2 \right )^{2} \right )}}dx

=\int_{0}^{4}\frac{1}{\sqrt-{\left \{ \left ( x-2 \right )^{2}-4 \right \}}}dx

                                                                            \left [ \because \left ( a-b \right )^{2}=a^{2}-2ab+b^{2} \right ]

=\int_{0}^{4} \frac{1}{\sqrt{4-(x-2)^{2}}} d x

=\int_{0}^{4} \frac{1}{\sqrt{2^{2}-(x-2)^{2}}} d x\left[\because \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\sin ^{-1}\left(\frac{x}{a}\right)\right]

=\left[\sin ^{-1}\left(\frac{x-2}{2}\right)\right]_{0}^{4}

=\left[\sin ^{-1}\left(\frac{4-2}{2}\right)-\sin ^{-1}\left(\frac{0-2}{2}\right)\right]

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

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

=\left[\sin ^{-1}(1)+\sin ^{-1}(1)\right]\left[\because \sin ^{-1}(-\theta)=-\sin \theta\right]

=2 \sin ^{-1}(1)

=2.\frac{\pi }{2}

=\pi

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