मराठी

Π / 2 ∫ 0 1 1 + Cot 7 X D X

Advertisements
Advertisements

प्रश्न

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

बेरीज
Advertisements

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 19: Definite Integrals - Revision Exercise [पृष्ठ १२२]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 19 Definite Integrals
Revision Exercise | Q 37 | पृष्ठ १२२

संबंधित प्रश्‍न

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

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

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

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

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

\[\int\limits_0^{\pi/2} \frac{\sin \theta}{\sqrt{1 + \cos \theta}} d\theta\]

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

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


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

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


Evaluate the following integral:

\[\int\limits_{- 2}^2 \left| 2x + 3 \right| dx\]

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

\[\int\limits_0^\pi x \sin^3 x\ dx\]

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

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

If f is an integrable function, show that

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

 


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

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

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

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

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

If \[\int\limits_0^a 3 x^2 dx = 8,\] write the value of a.

 

 


\[\int\limits_0^\sqrt{2} \left[ x^2 \right] dx .\]

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

\[\int\limits_1^\sqrt{3} \frac{1}{1 + x^2} dx\]  is equal to ______.

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

If \[\int\limits_0^a \frac{1}{1 + 4 x^2} dx = \frac{\pi}{8},\] then a equals

 


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

 


\[\int\limits_1^2 x\sqrt{3x - 2} dx\]


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


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


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


Evaluate the following using properties of definite integral:

`int_0^(i/2) (sin^7x)/(sin^7x + cos^7x)  "d"x`


Evaluate the following using properties of definite integral:

`int_0^1 x/((1 - x)^(3/4))  "d"x`


Evaluate the following:

Γ(4)


Evaluate the following integrals as the limit of the sum:

`int_0^1 (x + 4)  "d"x`


Evaluate the following integrals as the limit of the sum:

`int_1^3 (2x + 3)  "d"x`


Choose the correct alternative:

Γ(1) is


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


The value of `int_2^3 x/(x^2 + 1)`dx is ______.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×