Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int\limits_0^{\pi/2} \sin^2 x\ dx .\]
\[ = \int_0^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_0^\frac{\pi}{2} \left( 1 - \cos2x \right) dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{4}\]
APPEARS IN
संबंधित प्रश्न
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \left| \sin x - \cos x \right| dx\]
\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x - 1)/x^2 "d"x`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Evaluate the following:
`Γ (9/2)`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
