Mathematics

# Evaluate $\int \cot^{2}x\ dx$.

##### SOLUTION
$\int{\cot ^{2}x}\space dx$

We know that  $\text{cosec}^2x-\cot ^2x=1$
Therefore, $\Rightarrow \cot ^2x=\text{cosec}^2x-1$

$\Rightarrow \int{\cot ^{2}x}\space dx=\int{(\text{cosec}^2x-1)dx}$

$\Rightarrow \int{\text{cosec}^2x\space dx}-\int{1}\space dx$

This can be written as
$\Rightarrow -\int{-\text{cosec}^2x\space dx}-\int{1}\space dx$

This is a direct integral

$\Rightarrow -(\cot x)-x+C$

$\int{\cot ^{2}x}\space dx=-\cot x-x+C$

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

#### Realted Questions

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Solve:$\int _{ 0 }^{ \log { 2 } }{ \cos { 2x } dx }$=

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1 Verified Answer | Published on 17th 09, 2020

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Q5 Passage Hard

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$\int u(x)\, v(x)dx\, =\, u(x)\, v_{1}(x)\, -\, u^{}(x)v_{2}(x)\, +\, u^{}(x)\, v_{3}(x)\, -\, .\, +\, (-1)^{n\, -\, 1}u^{n\, -\, 1}(x)v_{n}(x)\, -\, (-1)^{n\, -\, 1}$ $\int\, u^{n}(x)v_{n}(x)\, dx$ where $v_{1}(x)\, =\, \int v(x)dx,\, v_{2}(x)\, =\, \int v_{1}(x)\, dx\, ..\, v_{n}(x)\, =\, \int v_{n\, -\, 1}(x) dx$

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