Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \sin^2 x\ d x\]
\[ = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \frac{1 - \cos2x}{2} dx\]
\[ = \frac{1}{2} \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 1 - \cos2x \right)dx\]
\[ = \frac{1}{2} \left[ x - \frac{\sin2x}{2} \right]_{- \frac{\pi}{2}}^\frac{\pi}{2} \]
\[ = \frac{1}{2}\left( \frac{\pi}{2} - 0 + \frac{\pi}{2} - 0 \right)\]
\[ = \frac{\pi}{2}\]
APPEARS IN
संबंधित प्रश्न
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
Write the coefficient a, b, c of which the value of the integral
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
`int_0^(2a)f(x)dx`
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{2\pi} \cos^7 x dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Choose the correct alternative:
Γ(1) is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
