Advertisements
Advertisements
प्रश्न
Advertisements
उत्तर
\[Let\ I = \int_0^\frac{\pi}{2} x^2 \cos x d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \sin x d x\]
\[ \Rightarrow I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} - \left[ - 2x \cos x \right]_0^\frac{\pi}{2} + \int_0^\frac{\pi}{2} - 2 \cos x d x\]
\[ \Rightarrow I = \left[ x^2 \sin x \right]_0^\frac{\pi}{2} + \left[ 2x \cos x \right]_0^\frac{\pi}{2} - \left[ 2 \sin x \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi^2}{4} - 2\]
APPEARS IN
संबंधित प्रश्न
Evaluate the following integral:
Evaluate each of the following integral:
\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\cos x}{1 + \sin^2 x} dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]
\[\int\limits_0^{\pi/2} \frac{1}{2 \cos x + 4 \sin x} dx\]
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"^(-4x) x^4 "d"x`
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Evaluate \[\int_{1}^{3}(2x+1)\,dx\].
