मराठी

1 ∫ − 1 1 1 + X 2 D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- 1}^1 \frac{1}{1 + x^2} dx\]
Advertisements

उत्तर

\[Let\ I = \int_{- 1}^1 \frac{1}{1 + x^2} d x . Then, \]
\[I = \left[ \tan^{- 1} x \right]_{- 1}^1 \]
\[ \Rightarrow I = \tan^{- 1} 1 - \tan^{- 1} \left( - 1 \right)\]
\[ \Rightarrow I = \frac{\pi}{4} - \left( - \frac{\pi}{4} \right)\]
\[ \Rightarrow I = \frac{\pi}{2}\]

shaalaa.com
Definite Integrals
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 20: Definite Integrals - Exercise 20.1 [पृष्ठ १६]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 20 Definite Integrals
Exercise 20.1 | Q 7 | पृष्ठ १६

संबंधित प्रश्‍न

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

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

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

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

\[\int\limits_1^3 \frac{\log x}{\left( x + 1 \right)^2} dx\]

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

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

\[\int_0^\frac{\pi}{4} \frac{\sin x + \cos x}{3 + \sin2x}dx\]

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

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

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

\[\int\limits_1^4 f\left( x \right) dx, where f\left( x \right) = \begin{cases}7x + 3 & , & \text{if }1 \leq x \leq 3 \\ 8x & , & \text{if }3 \leq x \leq 4\end{cases}\]


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

If `f` is an integrable function such that f(2a − x) = f(x), then prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 2 \int\limits_0^a f\left( x \right) dx\]

 


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

\[\int\limits_1^2 \left( x^2 - 1 \right) dx\]

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

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

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

Evaluate each of the following integral:

\[\int_0^\frac{\pi}{2} e^x \left( \sin x - \cos x \right)dx\]

 


Solve each of the following integral:

\[\int_2^4 \frac{x}{x^2 + 1}dx\]

If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.


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

 


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

The value of \[\int\limits_0^\pi \frac{1}{5 + 3 \cos x} dx\] is

 


The value of \[\int\limits_0^{\pi/2} \log\left( \frac{4 + 3 \sin x}{4 + 3 \cos x} \right) dx\] is 

 


Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .


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


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


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


Using second fundamental theorem, evaluate the following:

`int_1^2 (x - 1)/x^2  "d"x`


Evaluate the following using properties of definite integral:

`int_(- pi/4)^(pi/4) x^3 cos^3 x  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following:

`Γ (9/2)`


Evaluate the following:

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


Choose the correct alternative:

Γ(1) is


Find `int x^2/(x^4 + 3x^2 + 2) "d"x`


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×