मराठी

Π / 2 ∫ 0 X 2 Cos 2 X D X

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

उत्तर

\[\int_0^\frac{\pi}{2} x^2 \cos2x d x\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} 2x \frac{\sin2x}{2}dx\]

\[ = \left[ x^2 \frac{sin2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} x \sin 2x dx\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} - \left[ - x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} + \left[ - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx \right]\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \int_0^\frac{\pi}{2} \frac{\cos2x}{2}dx\]

\[ = \left[ x^2 \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} + \left[ x\frac{\cos2x}{2} \right]_0^\frac{\pi}{2} - \frac{1}{2} \left[ \frac{\sin2x}{2} \right]_0^\frac{\pi}{2} \]

\[ = 0 - \frac{\pi}{4} - 0\]

\[ = \frac{- \pi}{4}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 21 | पृष्ठ १२१

संबंधित प्रश्‍न

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

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

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

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

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

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

\[\int_0^\frac{\pi}{4} \left( \tan x + \cot x \right)^{- 2} dx\]

\[\int\limits_1^2 \frac{3x}{9 x^2 - 1} dx\]

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

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

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

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

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

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

Evaluate each of the following integral:

\[\int_0^{2\pi} \log\left( \sec x + \tan x \right)dx\]

 


Evaluate each of the following integral:

\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]


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

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

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{a - \sin \theta}{a + \sin \theta} \right) d\theta\]

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

 


If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]


Write the coefficient abc of which the value of the integral

\[\int\limits_{- 3}^3 \left( a x^2 + bx + c \right) dx\] is independent.

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

The value of \[\int\limits_0^\pi \frac{x \tan x}{\sec x + \cos x} dx\] is __________ .


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


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


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


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


\[\int\limits_{- \pi/4}^{\pi/4} \left| \tan x \right| dx\]


\[\int\limits_0^\pi \frac{x}{a^2 - \cos^2 x} dx, a > 1\]


Using second fundamental theorem, evaluate the following:

`int_1^2 (x "d"x)/(x^2 + 1)`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

`Γ (9/2)`


If f(x) = `{{:(x^2"e"^(-2x)",", x ≥ 0),(0",", "otherwise"):}`, then evaluate `int_0^oo "f"(x) "d"x`


Choose the correct alternative:

Γ(1) is


Which integral has lower and upper limits?


Evaluate \[\int_{0}^{1}x\,dx\].


If velocity changes with time, what does a definite integral give over a time interval?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×