मराठी

1 ∫ 0 √ 1 − X 1 + X D X

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प्रश्न

\[\int\limits_0^1 \sqrt{\frac{1 - x}{1 + x}} dx\]
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उत्तर

\[Let\ I = \int_0^1 \sqrt{\frac{1 - x}{1 + x}} d\ x . Then, \]
\[I = \int_0^1 \sqrt{\frac{1 - x}{1 + x}} \times \frac{\sqrt{1 - x}}{\sqrt{1 - x}} d x\]
\[ \Rightarrow I = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \int_0^1 \frac{1}{\sqrt{1 - x^2}} dx - \int_0^1 \frac{x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \left[ \sin^{- 1} x \right]_0^1 + \frac{1}{2} \int_0^1 \frac{- 2x}{\sqrt{1 - x^2}} dx\]
\[ \Rightarrow I = \left[ \sin^{- 1} x \right]_0^1 + \frac{1}{2} \left[ 2\sqrt{1 - x^2} \right]_0^1 \]
\[ \Rightarrow I = \frac{\pi}{2} - 0 + 0 - 1\]
\[ \Rightarrow I = \frac{\pi}{2} - 1\]

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पाठ 19: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.2 | Q 37 | पृष्ठ ३९

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