Advertisements
Advertisements
प्रश्न
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 the following integral:
Evaluate each of the following integral:
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.
Evaluate :
\[\int\limits_1^2 \frac{1}{x^2} e^{- 1/x} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_0^a \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
`int_(-1)^1 x^3 "e"^(x^4) "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Integrate `((2"a")/sqrt(x) - "b"/x^2 + 3"c"root(3)(x^2))` w.r.t. x
Evaluate the following:
`int ((x^2 + 2))/(x + 1) "d"x`
Find: `int logx/(1 + log x)^2 dx`