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
If \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]
If f(2a − x) = −f(x), prove that
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
\[\int\limits_0^\pi \frac{1}{1 + \sin x} dx\] equals
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^3 \left( x^2 + 1 \right) dx\]
Evaluate the following using properties of definite integral:
`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
The value of `int_2^3 x/(x^2 + 1)`dx is ______.
Which notation represents a definite integral?
Using \[F\] as an antiderivative, how is \[\int_{a}^{b}f(x)\,dx\] evaluated?
