Let 2x=t ⇒ #160; 2dx=dt ⇒ I = 12∫ tan^2tdt=12∫ (sec^2t 1)dt=12∫ sec^2tdt 12∫ dt =tant2 t2+c=tan(2x) 2x2+C . ...

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Evaluate: ∫t...

Question

Evaluate:
$\int {tan}^{2} 2 x d x$

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Solution

Let $2 x = t$

$\Rightarrow 2 d x = d t$

$\Rightarrow I =$ $\frac{1}{2} \int t a n^{2} t d t = \frac{1}{2} \int (s e c^{2} t - 1) d t = \frac{1}{2} \int s e c^{2} t d t - \frac{1}{2} \int d t$

$= \frac{t a n t}{2} - \frac{t}{2} + c = \frac{t a n (2 x) - 2 x}{2} + C$ .

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