Mathematics

# Solve :$\displaystyle\int { \tan ^{ 2 }{ \left( 2x-3 \right) } } dx$

##### SOLUTION
$\displaystyle\int{{\tan}^{2}{\left(2x-3\right)}dx}$

Let $t=2x-3\Rightarrow\,dt=2\,dx$

$=\dfrac{1}{2}\displaystyle\int{{\tan}^{2}{t}\ dt}$

$=\dfrac{1}{2}\displaystyle\int{\left({\sec}^{2}{t}-1\right)dt}$

$=\dfrac{1}{2}\left[\tan{t}-t\right]+c$ where $\displaystyle\int{{\sec}^{2}{x}dx}=\tan{x}+c$ and $\displaystyle\int{dx}=x+c$

$=\dfrac{1}{2}\left[\tan{\left(2x-3\right)}-2x-3\right]+c$ where $t=2x-3$

$=\dfrac{1}{2}\tan{\left(2x-3\right)}-x-\dfrac{3}{2}+c$

$=\dfrac{1}{2}\tan{\left(2x-3\right)}-x+C$ where $-\dfrac{3}{2}+c=C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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