हिंदी

Π / 4 ∫ 0 E X Sin X D X

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[Let, I = \int_0^\frac{\pi}{4} e^x \sin x d x ..............(1)\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \int_0^\frac{\pi}{4} e^x \cos x dx\]
\[ = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - \int_0^\frac{\pi}{4} e^x \sin x dx\]
\[ \Rightarrow I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} - I ..............\left[\text{Using (1)} \right] \]
\[ \Rightarrow 2I = \left[ - e^x \cos x \right]_0^\frac{\pi}{4} + \left[ e^x \sin x \right]_0^\frac{\pi}{4} \]
\[ = - \frac{1}{\sqrt{2}} e^\frac{\pi}{4} + 1 + \frac{1}{\sqrt{2}} e^\frac{\pi}{4} - 0\]
\[ = 1\]
\[\text{Hence }I = \frac{1}{2}\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 19 Definite Integrals
Revision Exercise | Q 27 | पृष्ठ १२२

संबंधित प्रश्न

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

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

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \frac{1}{1 + \sin x} dx\]

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

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

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

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

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

 


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

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

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

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

\[\int\limits_1^4 \left( x^2 - x \right) dx\]

\[\int\limits_a^b x\ dx\]

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

\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

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 \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


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

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

\[\int\limits_0^{\pi/2} \sin\ 2x\ \log\ \tan x\ dx\]  is equal to 

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

Evaluate : \[\int e^{2x} \cdot \sin \left( 3x + 1 \right) dx\] .


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


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


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


\[\int\limits_0^{15} \left[ x^2 \right] dx\]


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


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_0^(1/4) sqrt(1 - 4)  "d"x`


Using second fundamental theorem, evaluate the following:

`int_0^1 x"e"^(x^2)  "d"x`


Evaluate the following:

`int_0^oo "e"^(-4x) x^4  "d"x`


Choose the correct alternative:

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


Choose the correct alternative:

`int_0^oo x^4"e"^-x  "d"x` is


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×