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
\[\int\limits_0^{\pi/2} \frac{1}{2 + \cos x} dx\] equals
If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\] then the value of I10 + 90I8 is
The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]
\[\int\limits_2^3 e^{- x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^(1/4) sqrt(1 - 4) "d"x`
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Evaluate the following:
`int_1^4` f(x) dx where f(x) = `{{:(4x + 3",", 1 ≤ x ≤ 2),(3x + 5",", 2 < x ≤ 4):}`
Find `int x^2/(x^4 + 3x^2 + 2) "d"x`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
