मराठी

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

Advertisements
Advertisements

प्रश्न

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

उत्तर

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

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.1 | Q 51 | पृष्ठ १७

संबंधित प्रश्‍न

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

\[\int\limits_1^e \frac{\log x}{x} dx\]

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

\[\int\limits_0^\pi \left( \sin^2 \frac{x}{2} - \cos^2 \frac{x}{2} \right) dx\]

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

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

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

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

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

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

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{1 + \sqrt{\tan x}} dx\]

\[\int\limits_0^{\pi/2} \frac{1}{1 + \tan x}\]

 


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

\[\int\limits_0^\pi \frac{x \tan x}{\sec x \ cosec x} dx\]

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

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

\[\int\limits_{- \pi/2}^{\pi/2} \sin^3 x\ dx .\]

\[\int\limits_0^\infty e^{- x} dx .\]

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

Evaluate each of the following integral:

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

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

The value of \[\int\limits_0^1 \tan^{- 1} \left( \frac{2x - 1}{1 + x - x^2} \right) dx,\] is


The value of \[\int\limits_{- \pi/2}^{\pi/2} \left( x^3 + x \cos x + \tan^5 x + 1 \right) dx, \] is 


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


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


\[\int\limits_1^3 \left| x^2 - 2x \right| dx\]


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


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


\[\int\limits_0^\pi \frac{x \tan x}{\sec x + \tan x} dx\]


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×