English

Π / 4 ∫ π / 3 ( Tan X + Cot X ) 2 D X

Advertisements
Advertisements

Question

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

Solution

\[Let\ I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan x + \cot x \right)^2 d x . Then, \]
\[I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan^2 x + \cot^2 x + 2 \tan x \cot x \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \tan^2 x + \cot^2 x + 2 \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \sec^2 x - 1 + {cosec}^2 x - 1 + 2 \right) dx\]
\[ \Rightarrow I = \int_\frac{\pi}{3}^\frac{\pi}{4} \left( \sec^2 x + {cosec}^2 x \right) dx\]
\[ \Rightarrow I = \left[ \tan x - \cot x \right]_\frac{\pi}{3}^\frac{\pi}{4} \]
\[ \Rightarrow I = \left( 1 - 1 \right) - \left( \sqrt{3} - \frac{1}{\sqrt{3}} \right)\]
\[ \Rightarrow I = \frac{- 2}{\sqrt{3}}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 19: Definite Integrals - Exercise 20.1 [Page 16]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 19 Definite Integrals
Exercise 20.1 | Q 21 | Page 16

RELATED QUESTIONS

\[\int\limits_4^9 \frac{1}{\sqrt{x}} dx\]

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

\[\int\limits_0^1 \frac{2x + 3}{5 x^2 + 1} dx\]

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

\[\int\limits_0^2 \frac{1}{\sqrt{3 + 2x - x^2}} dx\]

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

\[\int\limits_0^{\pi/2} \frac{1}{5 \cos x + 3 \sin x} dx\]

\[\int\limits_0^{\pi/4} \frac{\tan^3 x}{1 + \cos 2x} dx\]

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

\[\int\limits_0^1 \frac{24 x^3}{\left( 1 + x^2 \right)^4} dx\]

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

\[\int_0^\frac{\pi}{4} \frac{\sin^2 x \cos^2 x}{\left( \sin^3 x + \cos^3 x \right)^2}dx\]

Evaluate the following integral:

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

\[\int\limits_0^5 \frac{\sqrt[4]{x + 4}}{\sqrt[4]{x + 4} + \sqrt[4]{9 - x}} dx\]

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

Evaluate the following integral:

\[\int_{- 1}^1 \left| xcos\pi x \right|dx\]

 


If `f` is an integrable function such that f(2a − x) = f(x), then prove that

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

 


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

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

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

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

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

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

The value of the integral \[\int\limits_0^\infty \frac{x}{\left( 1 + x \right)\left( 1 + x^2 \right)} dx\]

 


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

\[\int\limits_{\pi/6}^{\pi/3} \frac{1}{\sin 2x} dx\]  is equal to

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

 


Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .


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


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


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


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


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


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


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


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


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


Using second fundamental theorem, evaluate the following:

`int_0^(pi/2) sqrt(1 + cos x)  "d"x`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×