Mathematics

# Integrate the following:$\displaystyle \int\limits_{\pi /3}^{\pi /2} {{{(\tan x + \cot x)}^2}dx}$

##### SOLUTION
consider, $I = \displaystyle \int_{\pi/3}^{\pi/2} (\tan x + \cot x)^2dx$

$I = \displaystyle \int_{\pi/3}^{\pi/2} (\tan^2 x +2\tan x. \cot x+\cot^2x)dx$

$I = \displaystyle \int_{\pi/3}^{\pi/2} (\sec^2-1+2\times 1 + cosec^2 x-1)dx$

$I = \displaystyle \int_{\pi/3}^{\pi/2}\sec^2x\, dx + \int_{\pi/3}^{\pi/2}cesec^2x\, dx$

$=[\tan x - \cot x]^{\pi/2}_{\pi/3}$

$= -\sqrt{3} - \left(-\dfrac{1}{\sqrt{3}}\right) = -\sqrt{3} + \dfrac{1}{\sqrt{3}} = \dfrac{-2\sqrt{3}}{3}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

#### Realted Questions

Q1 Single Correct Hard
If $\displaystyle I_{1} = \int_{0}^{\frac{\pi}2} f(\sin 2x) \sin x\>dx$
and $\displaystyle I_{2} = \int_{0}^{\frac{\pi}4} f(\cos 2x) \cos x\>dx,$ then $\displaystyle \frac{I_{1}}{I_{2}}$ is equal to
• A. $1$
• B. $\dfrac{1}{\sqrt{2}}$
• C. $2$
• D. $\sqrt{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Single Correct Hard
The value of $\displaystyle\int{\frac{dx}{{(x+a)}^{\displaystyle\frac{8}{7}}{(x-b)}^{\displaystyle\frac{6}{7}}}}$, is equal to
• A. $\displaystyle\frac{3}{2(a+b)}{\left(\frac{x+a}{x-b}\right)}^{\displaystyle\frac{2}{3}}+C$
• B. $\displaystyle\frac{3}{(a+b)}{\left(\frac{x-b}{x+a}\right)}^{\displaystyle\frac{1}{3}}+C$
• C. None of these
• D. $\displaystyle\frac{7}{(a+b)}{\left(\frac{x-b}{x+a}\right)}^{\displaystyle\frac{1}{7}}+C$

1 Verified Answer | Published on 17th 09, 2020

Q3 Subjective Medium
Prove that:
$\displaystyle \int \sqrt{x^2+a^2}dx=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\log |x+\sqrt{x^2+a^2}|+c$

1 Verified Answer | Published on 17th 09, 2020

Q4 One Word Medium
Prove that $\displaystyle \int\frac{e^{\log \left ( 1+1/x^{2} \right )}}{x^{2}+1/x^{2}}dx=\frac{1}{\sqrt{\left ( 2 \right )}}\tan ^{-1}\left ( x-\frac{1}{x} \right )$

Let $\displaystyle f\left ( x \right )=\frac{\sin 2x \cdot \sin \left ( \dfrac{\pi }{2}\cos x \right )}{2x-\pi }$