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प्रश्न
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उत्तर
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shaalaa.com
Definite Integrals
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संबंधित प्रश्न
\[\int\limits_1^e \frac{e^x}{x} \left( 1 + x \log x \right) dx\]
\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]
\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]
\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]
\[\int\limits_0^\pi x \log \sin x\ dx\]
\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
