मराठी

Evaluate Each of the Following Integral: ∫ π 4 0 Sin 2 X D X

Advertisements
Advertisements

प्रश्न

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \sin2xdx\]
बेरीज
Advertisements

उत्तर

\[\int_0^\frac{\pi}{4} \sin2xdx\]
\[ = \left.\frac{- \cos2x}{2}\right|_0^\frac{\pi}{4} \]
\[ = - \frac{1}{2}\left( \cos\frac{\pi}{2} - \cos0 \right)\]
\[ = - \frac{1}{2} \times \left( 0 - 1 \right)\]
\[ = \frac{1}{2}\]
shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Very Short Answers [पृष्ठ ११५]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Very Short Answers | Q 27 | पृष्ठ ११५

संबंधित प्रश्‍न

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

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

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

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

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

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

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

\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{3/2}} dx\]

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

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

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

\[\int\limits_0^\pi x \sin x \cos^4 x\ dx\]

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

\[\int\limits_0^3 \left( 2 x^2 + 3x + 5 \right) dx\]

If \[\int\limits_0^1 \left( 3 x^2 + 2x + k \right) dx = 0,\] find the value of k.

 


If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


Evaluate : 

\[\int\limits_2^3 3^x dx .\]

\[\int\limits_0^2 \left[ x \right] dx .\]

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


`int_0^1 sqrt((1 - "x")/(1 + "x")) "dx"`


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

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

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

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

 


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


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


\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]


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


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


\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]


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


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_1^2 (x "d"x)/(x^2 + 1)`


Evaluate the following:

`int_0^2 "f"(x)  "d"x` where f(x) = `{{:(3 - 2x - x^2",", x ≤ 1),(x^2 + 2x - 3",", 1 < x ≤ 2):}`


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int sqrt((1 + x)/(1 - x)) "d"x`, x ≠1


Verify the following:

`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×