English

2 π ∫ 0 E X / 2 Sin ( X 2 + π 4 ) D X - Mathematics

Advertisements
Advertisements

Question

\[\int\limits_0^{2\pi} e^{x/2} \sin\left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]
Advertisements

Solution

\[Let\ I = \int_0^{2\pi} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} - \frac{2}{2} e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) dx\]
\[\text{Again, integrating second term by parts}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left\{ \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - \int_0^{2\pi} \frac{2}{2} e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) d x \right\}\]
\[ \Rightarrow I = \left[ - 2 e^\frac{x}{2} \cos \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} + \left[ 2 e^\frac{x}{2} \sin \left( \frac{x}{2} + \frac{\pi}{4} \right) \right]_0^{2\pi} - I\]
\[ \Rightarrow 2I = \frac{2}{\sqrt{2}} e^\pi + \frac{2}{\sqrt{2}} - \frac{2}{\sqrt{2}} e^\pi - \frac{2}{\sqrt{2}} = 0\]
\[ \Rightarrow I = 0\]

shaalaa.com
Definite Integrals
  Is there an error in this question or solution?
Chapter 20: Definite Integrals - Exercise 20.1 [Page 17]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 20 Definite Integrals
Exercise 20.1 | Q 51 | Page 17

RELATED QUESTIONS

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]

\[\int\limits_0^{\pi/2} \cos^4\ x\ dx\]

 


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

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

\[\int\limits_{\pi/2}^\pi e^x \left( \frac{1 - \sin x}{1 - \cos x} \right) dx\]

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

\[\int\limits_0^a \sqrt{a^2 - x^2} dx\]

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

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

\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]

\[\int_0^\frac{\pi}{2} \frac{\tan x}{1 + m^2 \tan^2 x}dx\]

\[\int_0^\frac{1}{2} \frac{1}{\left( 1 + x^2 \right)\sqrt{1 - x^2}}dx\]

\[\int_0^2 2x\left[ x \right]dx\]

\[\int\limits_0^7 \frac{\sqrt[3]{x}}{\sqrt[3]{x} + \sqrt[3]{7} - x} dx\]

If  \[f\left( a + b - x \right) = f\left( x \right)\] , then prove that \[\int_a^b xf\left( x \right)dx = \frac{a + b}{2} \int_a^b f\left( x \right)dx\]

 


\[\int\limits_0^{\pi/2} \frac{\sqrt{\cot x}}{\sqrt{\cot x} + \sqrt{\tan x}} dx\]

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

 


\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]

If f (x) is a continuous function defined on [0, 2a]. Then, prove that

\[\int\limits_0^{2a} f\left( x \right) dx = \int\limits_0^a \left\{ f\left( x \right) + f\left( 2a - x \right) \right\} dx\]

 


\[\int\limits_0^2 \left( x + 3 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 4 \right) dx\]

\[\int\limits_0^2 \left( 3 x^2 - 2 \right) dx\]

\[\int\limits_0^2 \left( x^2 + 2x + 1 \right) dx\]

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

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

\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

\[\int\limits_{- 1}^1 x\left| x \right| dx .\]

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


\[\int\limits_0^1 \frac{x}{\left( 1 - x \right)^\frac{5}{4}} dx =\]

The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


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


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


\[\int\limits_0^1 \log\left( 1 + x \right) dx\]


\[\int\limits_0^1 \left| 2x - 1 \right| dx\]


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 using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Choose the correct alternative:

The value of `int_(- pi/2)^(pi/2) cos  x  "d"x` is


Choose the correct alternative:

`Γ(3/2)`


`int (cos2x - cos 2theta)/(cosx - costheta) "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×