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Algebra of Limits

Solve: ∫ ta...

Question

Solve:
$\int t a n^{3} 2 x sec 2 x d x =$

A

s e c^{3} 2 x + 3 sec 2 x + c

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B

\frac{1}{6} (s e c^{3} 2 x - 3 sec 2 x) + c

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C

s e c^{3} 2 x - 3 sec 2 x + c

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D

\frac{- 1}{6} (s e c^{3} 2 x - 3 sec 2 x) + c

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Solution

The correct option is B $\frac{1}{6} (s e c^{3} 2 x - 3 sec 2 x) + c$
$\int t a n^{3} 2 x s e c 2 x d x \cdot$

$\int (t a n 2 x s e c 2 x) \cdot (t a n^{2} 2 x) d x \cdot$

Let $s e c 2 x = t$

$2 s e c 2 x t a n 2 x \cdot d x = d t$

$s e c 2 x \cdot t a n 2 x \cdot d x = 1 / 2 d t \cdot$

$\int (t^{2} - 1)^{1} /_{2} d t \cdot$

$∵ t a n^{2} 2 x = 1 + s e c^{2} 2 x$

$^{1} /_{2} \int (1 + t^{2}) d t$

$\Rightarrow^{1} /_{2} [- t +^{t} t^{3} /_{3}] + c$

$\Rightarrow^{1} /_{2} [t^{3} /_{3} - t]_{+ c} =^{1} /_{2} [\frac{s e c^{3} 2 x}{3} - s e c 2 x] + c$

$= [\frac{s e c^{3} 2 x}{6} - \frac{s e c 2 x}{2} + c]$

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