Advertisements
Advertisements
प्रश्न
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Advertisements
उत्तर
We have
\[I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos x}{\left( 1 + e^x \right)}dx . . . . . \left( i \right)\]
\[\text{ Using property } \int_a^b f\left( x \right) dx = \int_a^b f\left( a + b - x \right) dx, \text { we get }\]
\[I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos\left( 0 - x \right)}{1 + e^\left( 0 - x \right)}dx\]
\[ = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \frac{\cos x}{1 + e^{- x}}dx\]
\[ \Rightarrow I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} e^x \frac{\left( \cos x \right)}{\left( 1 + e^x \right)}dx . . . . . \left( ii \right)\]
Adding (i) and (ii), we get
\[2I = \int\limits_\frac{- \pi}{2}^\frac{\pi}{2} \cos x dx = \left[ \sin x \right]_\frac{- \pi}{2}^\frac{\pi}{2} = 1 + 1 = 2\]
\[ \therefore I = 1\]
APPEARS IN
संबंधित प्रश्न
The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\] is
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
\[\int\limits_0^1 \cos^{- 1} x dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_1^2 \frac{x + 3}{x\left( x + 2 \right)} dx\]
\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]
\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]
\[\int\limits_2^3 e^{- x} dx\]
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following using properties of definite integral:
`int_(- pi/2)^(pi/2) sin^2theta "d"theta`
Evaluate the following integrals as the limit of the sum:
`int_1^3 (2x + 3) "d"x`
`int "e"^x ((1 - x)/(1 + x^2))^2 "d"x` is equal to ______.
