हिंदी

Π / 2 ∫ − π / 2 Log ( 2 − Sin X 2 + Sin X ) D X - Mathematics

Advertisements
Advertisements

प्रश्न

\[\int\limits_{- \pi/2}^{\pi/2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) dx\]
Advertisements

उत्तर

\[Let\ I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) d x\]
\[Here, f\left( x \right) = log\left( \frac{2 - \sin x}{2 + \sin x} \right)\]
\[f\left( - x \right) = log\left( \frac{2 - \sin\left( - x \right)}{2 + \sin\left( - x \right)} \right) = log\left( \frac{2 + \sin x}{2 - \sin x} \right) = - log\left( \frac{2 - \sin x}{2 + \sin x} \right) = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}\]
\[ \therefore I = 0\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Exercise 20.5 [पृष्ठ ९५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.5 | Q 28 | पृष्ठ ९५

संबंधित प्रश्न

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

\[\int\limits_0^{\pi/6} \cos x \cos 2x\ dx\]

\[\int_0^1 x\log\left( 1 + 2x \right)dx\]

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

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

\[\int\limits_0^{\pi/4} \left( \sqrt{\tan}x + \sqrt{\cot}x \right) dx\]

\[\int_0^\frac{\pi}{2} \frac{\cos^2 x}{1 + 3 \sin^2 x}dx\]

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

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

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

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

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

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

If f(2a − x) = −f(x), prove that

\[\int\limits_0^{2a} f\left( x \right) dx = 0 .\]

If f is an integrable function, show that

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


If f(x) is a continuous function defined on [−aa], then prove that 

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

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

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

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

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

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

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

 


\[\int\limits_0^\pi \cos^5 x\ dx .\]

Evaluate each of the following integral:

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

 


\[\int\limits_1^2 \log_e \left[ x \right] dx .\]

\[\int\limits_{- \pi/2}^{\pi/2} \sin\left| x \right| dx\]  is equal to

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

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


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


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


\[\int\limits_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{\left( 1 - \cos x \right)^{5/2}} dx\]


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


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


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:

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


Choose the correct alternative:

`int_0^1 (2x + 1)  "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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×