Mathematics

# $\int {(\tan x-\cot x)^2dx}=$

$\tan x-\cot x-4x+C$

##### SOLUTION
$I=\int { { \left( \tan { x } -\cot { x } \right) }^{ 2 } } dx=\int { \cfrac { { \left( \tan ^{ 2 }{ x } -1 \right) }^{ 2 } }{ \tan ^{ 2 }{ x } } } dx=\int { \cfrac { \tan ^{ 4 }{ x } +1-2\tan ^{ 2 }{ x } }{ \tan ^{ 2 }{ x } } } dx$
$=\int { \tan ^{ 2 }{ x } } dx+\int { \cot ^{ 2 }{ x } } dx-2\int { dx } =\int { \left[ \left( \tan ^{ 2 }{ x } +1 \right) -1 \right] } dx+\int { \cfrac { \cos ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } } dx-2x+C$
$=\int { \sec ^{ 2 }{ x } } dx-\int { dx } +\int { \cfrac { 1-\sin ^{ 2 }{ x } }{ \sin ^{ 2 }{ x } } } dx-2x+C$
$=\tan { x } -x+\int { co\sec ^{ 2 }{ x } } dx-\int { dx } -2x+C=\tan { x } -4x+\int { co\sec ^{ 2 }{ x } } dx+C=\tan { x } -4x-\cot { x } +C\quad$
$I=\tan { x } -\cot { x } -4x+C$

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

#### Realted Questions

Q1 Single Correct Hard
Let $f(x)$ be a positive function. Let
$I_{1} = \int_{1 - k}^{k} xf\left \{x(1 - x)\right \} dx$,
$I_{2} = \int_{1 - k}^{k} f\left \{x(1 - x) \right \} dx$,
where $2k - 1 > 0$, then $\dfrac {I_{1}}{I_{2}}$ is
• A. $2$
• B. $k$
• C. $1$
• D. $\dfrac {1}{2}$

1 Verified Answer | Published on 17th 09, 2020

Q2 Subjective Medium
Solve : $\displaystyle \int \dfrac{x^3 \, dx}{(x^2 + a^2)(x^2 + b^2)}$

1 Verified Answer | Published on 17th 09, 2020

Q3 Single Correct Hard

If the integral $\displaystyle \int\frac{5\tan x}{\tan x-2}dx=x+$ $a\ ln$ $|\sin x-2\cos \mathrm{x}|+k$ then $a$ is equal to:
• A. $1$
• B. $-2$
• C. $-1$
• D. $2$

1 Verified Answer | Published on 17th 09, 2020

Q4 Subjective Medium
Solve $\int(4x+2)\sqrt{{x}^{2}+x+1}dx$

Solve: $\int x - \sin^2 x \,dx$