मराठी

Π / 2 ∫ 0 X Cos X D X

Advertisements
Advertisements

प्रश्न

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

उत्तर

\[Let I = \int_0^\frac{\pi}{2}\ x\ \cos\ x\ d\ x\ . Then, \]
\[\text{Integrating by parts}\]
\[I = \left[ x \sin x \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 1 \sin x d x\]
\[ \Rightarrow I = \left[ x \sin x \right]_0^\frac{\pi}{2} + \left[ \cos x \right]_0^\frac{\pi}{2} \]
\[ \Rightarrow I = \frac{\pi}{2} - 1\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Exercise 20.1 [पृष्ठ १७]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Exercise 20.1 | Q 27 | पृष्ठ १७

संबंधित प्रश्‍न

\[\int\limits_2^3 \frac{x}{x^2 + 1} dx\]

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

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

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

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

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

\[\int\limits_0^{( \pi )^{2/3}} \sqrt{x} \cos^2 x^{3/2} dx\]


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

\[\int\limits_0^9 f\left( x \right) dx, where f\left( x \right) \begin{cases}\sin x & , & 0 \leq x \leq \pi/2 \\ 1 & , & \pi/2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9\end{cases}\]

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

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

 


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

 


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

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

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

\[\int\limits_0^5 \left( x + 1 \right) dx\]

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

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

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

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

If \[\left[ \cdot \right] and \left\{ \cdot \right\}\] denote respectively the greatest integer and fractional part functions respectively, evaluate the following integrals:

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

 


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

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


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


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


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


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


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


\[\int\limits_0^\pi \frac{dx}{6 - \cos x}dx\]


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


\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]


Evaluate the following using properties of definite integral:

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


Evaluate the following:

`int_0^oo "e"^(-mx) x^6 "d"x`


Choose the correct alternative:

Γ(n) is


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


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


`int x^3/(x + 1)` is equal to ______.


Given `int "e"^"x" (("x" - 1)/("x"^2)) "dx" = "e"^"x" "f"("x") + "c"`. Then f(x) satisfying the equation is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×