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प्रश्न
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
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उत्तर
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संबंधित प्रश्न
\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]
\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]
\[\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx\]
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \sin x\ dx\]
\[\int\limits_0^2 \left( x^2 + x \right) dx\]
The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
Find: `int logx/(1 + log x)^2 dx`
