English

∫ π 3 π 6 ( Tan X + Cot X ) 2 D X

Advertisements
Advertisements

Question

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]
Sum
Advertisements

Solution

\[\int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan x + \cot x \right)^2 dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \tan^2 x + \cot^2 x + 2\tan x\cot x \right)dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \left( \sec^2 x - 1 + {cosec}^2 x - 1 + 2 \right)dx\]
\[ = \int_\frac{\pi}{6}^\frac{\pi}{3} \sec^2 xdx + \int_\frac{\pi}{6}^\frac{\pi}{3} {cosec}^2 xdx\]

\[= \left.\tan x\right|_\frac{\pi}{6}^\frac{\pi}{3} + \left.\left( - \cot x \right)\right|_\frac{\pi}{6}^\frac{\pi}{3} \]
\[ = \left( \tan\frac{\pi}{3} - \tan\frac{\pi}{6} \right) - \left( \cot\frac{\pi}{3} - \cot\frac{\pi}{6} \right)\]
\[ = \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right) - \left( \frac{1}{\sqrt{3}} - \sqrt{3} \right)\]
\[ = 2\sqrt{3} - \frac{2}{\sqrt{3}}\]
\[ = \frac{4}{\sqrt{3}}\]

 

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.1 [Page 18]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.1 | Q 66 | Page 18

RELATED QUESTIONS

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

\[\int\limits_{\pi/3}^{\pi/4} \left( \tan x + \cot x \right)^2 dx\]

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

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

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

\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]

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

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

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

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

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

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

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

 


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

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

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

\[\int\limits_0^\pi \log\left( 1 - \cos x \right) dx\]

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

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

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

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

\[\int_0^\frac{\pi^2}{4} \frac{\sin\sqrt{x}}{\sqrt{x}} dx\] equals


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

If \[\int\limits_0^1 f\left( x \right) dx = 1, \int\limits_0^1 xf\left( x \right) dx = a, \int\limits_0^1 x^2 f\left( x \right) dx = a^2 , then \int\limits_0^1 \left( a - x \right)^2 f\left( x \right) dx\] equals


If \[I_{10} = \int\limits_0^{\pi/2} x^{10} \sin x\ dx,\]  then the value of I10 + 90I8 is

 


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


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


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


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


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


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


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


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


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


Evaluate the following using properties of definite integral:

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


`int "e"^x ((1 - x)/(1 + x^2))^2  "d"x` is equal to ______.


What is the result of a definite integral?


What is an antiderivative of \[x\]?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×