Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_0^\frac{\pi}{2} \sqrt{1 - \cos2x}\ dx\]
\[ = \int_0^\frac{\pi}{2}\sqrt{2 \sin^2 x}\ dx\]
\[ = \int_0^\frac{\pi}{2} \sqrt{2} \sin x\ dx\]
\[ = - \sqrt{2} \left[ \cos x \right]_0^\frac{\pi}{2} \]
\[ = - \left( 0 - \sqrt{2} \right)\]
\[ = \sqrt{2}\]
\[\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
Evaluate each of the following integral:
Write the coefficient a, b, c of which the value of the integral
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Choose the correct alternative:
`Γ(3/2)`
`int x^9/(4x^2 + 1)^6 "d"x` is equal to ______.
