Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let I = \int_0^\frac{\pi}{2}\ x\ \cos\ x\ d\ x\ . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ x \sin x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 1 \sin x d x\]
\[ \Rightarrow I = \left[ x \sin x \right]_0^\frac{\pi}{2} + \left[ \cos x \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{2} - 1\]
APPEARS IN
संबंधित प्रश्न
Evaluate each of the following integral:
If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:
If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_2^3 \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}} dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`int_0^oo "e"^(- x/2) x^5 "d"x`
Evaluate the following integrals as the limit of the sum:
`int_1^3 x "d"x`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
