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 each of the following integral:
\[\int\limits_0^{2a} f\left( x \right) dx\] is equal to
\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]
\[\int\limits_0^{15} \left[ x^2 \right] dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_1^2 (x "d"x)/(x^2 + 1)`
Evaluate the following:
`int_(-1)^1 "f"(x) "d"x` where f(x) = `{{:(x",", x ≥ 0),(-x",", x < 0):}`
Evaluate the following using properties of definite integral:
`int_0^1 log (1/x - 1) "d"x`
Evaluate the following integrals as the limit of the sum:
`int_0^1 (x + 4) "d"x`
