Advertisements
Advertisements
Question
The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is
Options
1
e − 1
0
− 1
Advertisements
Solution
e − 1
\[Let\, I = \int_0^\frac{\pi}{2} \cos x\ e^{\sin x}\ d x\]
\[Let\ \sin x = t, then\ \cos x dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 e^t dt\]
\[ = \left[ e^t \right]_0^1 \]
\[ = e - 1\]
APPEARS IN
RELATED QUESTIONS
If f (x) is a continuous function defined on [0, 2a]. Then, prove that
\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.
The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Using second fundamental theorem, evaluate the following:
`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
