हिंदी

1 ∫ 0 √ 1 − X 1 + X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग
Advertisements

उत्तर

\[\int_0^1 \sqrt{\frac{1 - x}{1 + x}} d x\]

\[ = \int_0^1 \sqrt{\frac{1 - x}{1 + x} \times \frac{1 - x}{1 - x}} d x\]

\[ = \int_0^1 \frac{1 - x}{\sqrt{1 - x^2}} d x\]

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

\[ = \left[ \sin^{- 1} x \right]_0^1 + \left[ \sqrt{1 - x^2} \right]_0^1 \]

\[ = \frac{\pi}{2} - 1\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Revision Exercise [पृष्ठ १२१]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Revision Exercise | Q 17 | पृष्ठ १२१

संबंधित प्रश्न

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

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

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

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

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

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


\[\int\limits_0^{\pi/2} \sin 2x \tan^{- 1} \left( \sin x \right) dx\]

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

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

\[\int\limits_0^3 \left( x + 4 \right) dx\]

\[\int\limits_1^2 x^2 dx\]

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

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

Solve each of the following integral:

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

The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is 


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


Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]


\[\int\limits_1^e \log x\ dx =\]

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

 


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

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

 


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


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


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


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


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


Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`


Using second fundamental theorem, evaluate the following:

`int_0^1 "e"^(2x)  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 log (1/x - 1)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_0^1 x^2  "d"x`


Choose the correct alternative:

`int_0^1 (2x + 1)  "d"x` is


Choose the correct alternative:

`int_0^oo "e"^(-2x)  "d"x` is


Choose the correct alternative:

If f(x) is a continuous function and a < c < b, then `int_"a"^"c" f(x)  "d"x + int_"c"^"b" f(x)  "d"x` is


Find `int sqrt(10 - 4x + 4x^2)  "d"x`


Evaluate `int (x^2"d"x)/(x^4 + x^2 - 2)`


Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×