1 ∫ 0 Tan − 1 X 1 + X 2 D X

प्रश्न

\[\int\limits_0^1 \frac{\tan^{- 1} x}{1 + x^2} dx\]

उत्तर

\[Let\ I = \int_0^1 \frac{\tan^{- 1} x}{1 + x^2} d\ x . Then, \]
\[Let\ \tan^{- 1} x = t . Then, \frac{1}{1 + x^2} dx = dt\]
\[When\ x = 0, t = 0\ and\ x = 1, t = \frac{\pi}{4}\]
\[ \therefore I = \int_0^\frac{\pi}{4} t dt\]
\[ \Rightarrow I = \left[ \frac{t^2}{2} \right]_0^\frac{\pi}{4} \]
\[ \Rightarrow I = \frac{\pi^2}{32}\]

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Definite Integrals

क्या इस प्रश्न या उत्तर में कोई त्रुटि है?

Q 30 Q 29 Q 31

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

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12

अध्याय 19 Definite Integrals
Exercise 20.2 | Q 30 | पृष्ठ ३९

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