English

Π / 2 ∫ 0 1 1 + √ Tan X D X

Advertisements
Advertisements

Question

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

Solution

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

\[ = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{\tan\left( \frac{\pi}{2} - x \right)}} dx.....................\left[\text{Using }\int_0^a f\left( x \right) dx = \int_0^a f\left( a - x \right) dx\right]\]

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

\[\text{Adding (1) and (2) we get}\]

\[2I = \int_0^\frac{\pi}{2} \frac{1}{1 + \sqrt{\tan x}} + \frac{1}{1 + \sqrt{cotx}} d x\]

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

\[ = \int_0^\frac{\pi}{2} \frac{1 + \sqrt{cotx} + 1 + \sqrt{\tan x}}{1 + \sqrt{cotx} + \sqrt{\tan x} + \sqrt{\tan x \ cotx}} dx\]

\[ = \int_0^\frac{\pi}{2} \frac{2 + \sqrt{cotx} + \sqrt{\tan x}}{2 + \sqrt{cotx} + \sqrt{\tan x}} dx\]

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

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

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

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

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.5 [Page 94]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.5 | Q 6 | Page 94

RELATED QUESTIONS

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

\[\int\limits_{\pi/6}^{\pi/4} cosec\ x\ dx\]

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

\[\int\limits_0^{\pi/2} \sqrt{\sin \phi} \cos^5 \phi\ d\phi\]

 


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

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

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

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

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

Evaluate the following integral:

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

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

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

 


If f(2a − x) = −f(x), prove that

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

\[\int\limits_2^3 \left( 2 x^2 + 1 \right) dx\]

\[\int\limits_0^1 \left( 3 x^2 + 5x \right) dx\]

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

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

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

\[\int\limits_0^{\pi/2} \log \left( \frac{3 + 5 \cos x}{3 + 5 \sin x} \right) dx .\]

 


Evaluate each of the following integral:

\[\int_e^{e^2} \frac{1}{x\log x}dx\]

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

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

\[\lim_{n \to \infty} \left\{ \frac{1}{2n + 1} + \frac{1}{2n + 2} + . . . + \frac{1}{2n + n} \right\}\] is equal to

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

\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]


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


\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\left( 1 + \cos x \right)^2} dx\]


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


\[\int\limits_0^{\pi/4} \sin 2x \sin 3x dx\]


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


\[\int\limits_0^1 \left| \sin 2\pi x \right| dx\]


\[\int\limits_1^3 \left| x^2 - 4 \right| dx\]


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


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


Evaluate the following using properties of definite integral:

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


Evaluate the following 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


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


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×