हिंदी

2 π ∫ 0 E X / 2 Sin ( X 2 + π 4 ) D X

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
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Exercise 20.1 | Q 51 | पृष्ठ १७

संबंधित प्रश्न

\[\int\limits_{- 2}^3 \frac{1}{x + 7} dx\]

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

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

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

 


\[\int\limits_0^{\pi/2} \left( a^2 \cos^2 x + b^2 \sin^2 x \right) dx\]

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

Evaluate the following definite integrals:

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

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

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

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

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

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

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

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

\[\int\limits_0^a \sin^{- 1} \sqrt{\frac{x}{a + x}} dx\]

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

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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

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

If f is an integrable function, show that

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


\[\int\limits_3^5 \left( 2 - x \right) dx\]

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

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

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

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

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

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

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


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

\[\int\limits_0^{\pi/3} \frac{\cos x}{3 + 4 \sin x} dx\]


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


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


\[\int\limits_0^4 x dx\]


Find : `∫_a^b logx/x` dx


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×