हिंदी

Π / 2 ∫ 0 1 1 + Cot 7 X D X - Mathematics

Advertisements
Advertisements

प्रश्न

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

योग

उत्तर

\[Let, I = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 x} d x ..............(1)\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 \left( \frac{\pi}{2} - x \right)} d x\]

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \tan^7 x} d x ..............(2)\]

Adding (1) and (2)

\[2I = \int_0^\frac{\pi}{2} \frac{1}{1 + co t^7 x} + \frac{1}{1 + \tan^7 x} d x \]

\[ = \int_0^\frac{\pi}{2} \frac{2 + co t^7 x + \tan^7 x}{\left( 1 + co t^7 x \right)\left( 1 + \tan^7 x \right)}dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + co t^7 x + \tan^7 x}{2 + co t^7 x + \tan^7 x}dx\]

\[ = \int_0^\frac{\pi}{2} dx\]

\[ = \left[ x \right]_0^\frac{\pi}{2} \]

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

\[Hence, I = \frac{\pi}{4}\]

shaalaa.com
Definite Integrals
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 20: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Revision Exercise | Q 37 | पृष्ठ १२२

संबंधित प्रश्न

\[\int\limits_{\pi/4}^{\pi/2} \cot x\ dx\]


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

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

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

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

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

\[\int\limits_2^4 \frac{x}{x^2 + 1} dx\]

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

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

\[\int\limits_4^{12} x \left( x - 4 \right)^{1/3} dx\]

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

Evaluate the following integral:

\[\int\limits_{- 3}^3 \left| x + 1 \right| dx\]

\[\int_{- 1}^2 \left( \left| x + 1 \right| + \left| x \right| + \left| x - 1 \right| \right)dx\]

 


\[\int_{- \frac{\pi}{2}}^\pi \sin^{- 1} \left( \sin x \right)dx\]

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

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

Evaluate the following integral:

\[\int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]

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

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

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

\[\int\limits_0^{\pi/2} \sqrt{1 - \cos 2x}\ dx .\]

\[\int\limits_a^b \frac{f\left( x \right)}{f\left( x \right) + f\left( a + b - x \right)} dx .\]

\[\int\limits_0^2 \sqrt{4 - x^2} dx\]

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


\[\int\limits_0^1 e^\left\{ x \right\} dx .\]

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

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


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


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


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


\[\int\limits_0^\pi \cos 2x \log \sin x dx\]


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


\[\int\limits_2^3 e^{- x} dx\]


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

`int_(-1)^1 x^3 "e"^(x^4)  "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 (3"a"x)/("b"^2 + "c"^2x^2) "d"x`


Evaluate `int (x^2 + x)/(x^4 - 9) "d"x`


Verify the following:

`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×