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1 ∫ 0 Tan − 1 X D X

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प्रश्न

\[\int\limits_0^1 \tan^{- 1} x dx\]

योग
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उत्तर

\[\int_0^1 \tan^{- 1} x d x\]

\[ = \int_0^1 \tan^{- 1} x \times 1 d x\]

\[ = \left[ \tan^{- 1} x x \right]_0^1 - \int_0^1 \frac{x}{1 + x^2}dx\]

\[ = \left[ x \tan^{- 1} x \right]_0^1 - \frac{1}{2} \left[ \log\left( 1 + x^2 \right) \right]_0^1 \]

\[ = \frac{\pi}{4} - 0 - \frac{1}{2}\log2 + 0\]

\[ = \frac{\pi}{4} - \frac{1}{2}\log2\]

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अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 5 | पृष्ठ १२१

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