Advertisements
Advertisements
प्रश्न
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
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
Evaluate the following definite integrals:
If f(2a − x) = −f(x), prove that
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of the integral \[\int\limits_{- 2}^2 \left| 1 - x^2 \right| dx\] is ________ .
The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is
Evaluate : \[\int\limits_0^\pi \frac{x}{1 + \sin \alpha \sin x}dx\] .
\[\int\limits_0^1 \left| 2x - 1 \right| dx\]
\[\int\limits_{- \pi/2}^{\pi/2} \sin^9 x dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
Choose the correct alternative:
`Γ(3/2)`
Choose the correct alternative:
`int_0^oo x^4"e"^-x "d"x` is
`int (x + 3)/(x + 4)^2 "e"^x "d"x` = ______.
