English

The Value of π / 2 ∫ 0 Cos X E Sin X D X Is, (A) 1 (B) E − 1 (C) 0 (D) − 1

Advertisements
Advertisements

Question

The value of \[\int\limits_0^{\pi/2} \cos x\ e^{\sin x}\ dx\] is

 

Options

  •  1

  • e − 1

  • 0

  • − 1 

MCQ
Advertisements

Solution

e − 1

\[Let\, I = \int_0^\frac{\pi}{2} \cos x\ e^{\sin x}\ d x\]
\[Let\ \sin x = t, then\ \cos x dx = dt\]
\[When\ x = 0, t = 0\ and\ x = \frac{\pi}{2}, t = 1\]
\[\text{Therefore the integral becomes}\]
\[I = \int_0^1 e^t dt\]
\[ = \left[ e^t \right]_0^1 \]
\[ = e - 1\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - MCQ [Page 118]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
MCQ | Q 20 | Page 118

RELATED QUESTIONS

\[\int\limits_0^\infty \frac{1}{a^2 + b^2 x^2} dx\]

\[\int\limits_0^{\pi/2} \sqrt{1 + \cos x}\ dx\]

\[\int\limits_0^{\pi/2} x^2 \cos^2 x\ dx\]

\[\int\limits_1^2 \log\ x\ dx\]

\[\int\limits_1^2 \frac{x + 3}{x \left( x + 2 \right)} dx\]

\[\int\limits_0^1 x \left( 1 - x \right)^5 dx\]

\[\int\limits_1^2 \left( \frac{x - 1}{x^2} \right) e^x dx\]

\[\int_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\]

\[\int\limits_0^1 \frac{2x}{1 + x^4} dx\]

\[\int\limits_0^\pi \frac{1}{3 + 2 \sin x + \cos x} dx\]

\[\int\limits_0^\pi 5 \left( 5 - 4 \cos \theta \right)^{1/4} \sin \theta\ d \theta\]

\[\int\limits_0^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x} dx\]

 


\[\int\limits_0^\pi x \sin^3 x\ dx\]

\[\int\limits_0^{\pi/2} \cos x\ dx\]

\[\int\limits_0^2 \left( x^2 + 2 \right) dx\]

\[\int\limits_{- \pi/2}^{\pi/2} x \cos^2 x\ dx .\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

The value of the integral \[\int\limits_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx\]  is 


\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


\[\int\limits_0^\pi \frac{1}{a + b \cos x} dx =\]

Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


\[\int\limits_0^3 \frac{3x + 1}{x^2 + 9} dx =\]

The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


\[\int\limits_{- 1}^1 \left| 1 - x \right| dx\]  is equal to

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 

 


\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]


\[\int\limits_0^\pi \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\]


\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


\[\int\limits_0^{\pi/2} \frac{x}{\sin^2 x + \cos^2 x} dx\]


\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]


Using second fundamental theorem, evaluate the following:

`int_1^2 (x "d"x)/(x^2 + 1)`


Using second fundamental theorem, evaluate the following:

`int_0^3 ("e"^x "d"x)/(1 + "e"^x)`


Evaluate the following:

f(x) = `{{:("c"x",", 0 < x < 1),(0",",  "otherwise"):}` Find 'c" if `int_0^1 "f"(x)  "d"x` = 2


Evaluate the following:

`int_0^oo "e"^(- x/2) x^5  "d"x`


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


If x = `int_0^y "dt"/sqrt(1 + 9"t"^2)` and `("d"^2y)/("d"x^2)` = ay, then a equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×