Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\pi x \sin^3 x\ d x . . . (i)\]
\[ = \int_0^\pi \left( \pi - x \right) \sin^3 \left( \pi - x \right) d x\]
\[ = \int_0^\pi \left( \pi - x \right) \sin^3 x dx . . . (ii)\]
\[\text{Adding (i) and (ii) we get}\]
\[2I = \int_0^\pi \left( x + \pi - x \right) \sin^3 x\ d x\]
\[ = \int_0^\pi \pi \sin^3 x d x\]
\[ = \int_0^\pi \pi \frac{3 \sin x - \sin 3x}{4} d\ x\]
\[ = \frac{\pi}{4} \int_0^\pi \left( 3 \sin x - \sin 3x \right) d x\]
\[ = \frac{\pi}{4} \left[ - 3 \cos x + \frac{\cos 3x}{3} \right]_0^\pi \]
\[ = \frac{\pi}{4}\left[ - 3 \cos \pi + 3\cos 0 + \frac{\cos 3\pi}{3} - \frac{\cos 0}{3} \right]\]
\[ = \frac{\pi}{4}\left[ 3 + 3 + \frac{- 1}{3} - \frac{1}{3} \right]\]
\[ = \frac{\pi}{2}\left[ 3 - \frac{1}{3} \right]\]
\[ = \frac{\pi}{2} \times \frac{8}{3}\]
\[ = \frac{4\pi}{3}\]
\[ \therefore I = \frac{2\pi}{3}\]
APPEARS IN
संबंधित प्रश्न
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]
If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
\[\int\limits_0^1 \cos^{- 1} \left( \frac{1 - x^2}{1 + x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan^3 x} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
Choose the correct alternative:
Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is
