मराठी

1 ∫ 0 D D X { Sin − 1 ( 2 X 1 + X 2 ) } D X is Equal To(A) 0 (B) π (C) π/2 (D) π/4

Advertisements
Advertisements

प्रश्न

\[\int\limits_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} dx\] is equal to

पर्याय

  •  0

  •  π

  • π/2

  • π/4

MCQ
Advertisements

उत्तर

π/2

\[\text{We have}, \]

\[I = \int_0^1 \frac{d}{dx}\left\{ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\}dx\]

\[\text{We know since} \int f'(x) = f(x)\]

\[f(x) = si n^{- 1} \left( \frac{2x}{1 + x^2} \right) and f'(x) = \frac{d}{dx}\left\{ si n^{- 1} \left( \frac{2x}{1 + x^2} \right) \right\} \]

\[\text{Therefore}, I = \left[ \sin^{- 1} \left( \frac{2x}{1 + x^2} \right) \right]_0^1 \]

\[ = \sin^{- 1} \left( 1 \right) - \sin^{- 1} \left( 0 \right)\]

\[ = \frac{\pi}{2}\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - MCQ [पृष्ठ १२०]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
MCQ | Q 33 | पृष्ठ १२०

संबंधित प्रश्‍न

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

\[\int\limits_0^\infty e^{- x} dx\]

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 + 4 \sin x} dx\]

\[\int\limits_1^4 f\left( x \right) dx, where\ f\left( x \right) = \begin{cases}4x + 3 & , & \text{if }1 \leq x \leq 2 \\3x + 5 & , & \text{if }2 \leq x \leq 4\end{cases}\]

 


\[\int_0^\pi \cos x\left| \cos x \right|dx\]

\[\int_{- \frac{\pi}{2}}^\frac{\pi}{2} \left( 2\sin\left| x \right| + \cos\left| x \right| \right)dx\]

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

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

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

 


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

\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx, 0 < \alpha < \pi\]

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

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

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

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

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

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

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

 


\[\int\limits_{- 2}^1 \frac{\left| x \right|}{x} dx .\]

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{4} \tan\ xdx\]

 


\[\int\limits_0^1 \left\{ x \right\} dx,\] where {x} denotes the fractional part of x.  

 

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


The value of \[\int\limits_{- \pi}^\pi \sin^3 x \cos^2 x\ dx\] is 

 


The derivative of \[f\left( x \right) = \int\limits_{x^2}^{x^3} \frac{1}{\log_e t} dt, \left( x > 0 \right),\] is

 


If f (a + b − x) = f (x), then \[\int\limits_a^b\] x f (x) dx is equal to


\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


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


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


\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]


Choose the correct alternative:

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


Choose the correct alternative:

Using the factorial representation of the gamma function, which of the following is the solution for the gamma function Γ(n) when n = 8 is


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


What is an antiderivative of \[2x+1\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×