The correct option is A 1/6 sec^3 2x 1/2sec 2x+c∫ tan^3 2x sec 2x dx=∫ (sec^2 2x 1)sec 2x tan 2x dx =∫ (sec^3 2x tan 2x sec 2x tan 2x)dx =∫ sec^3 2x tan 2x ...

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Trigonometric Ratios of Multiples of an Angle

∫tan 3 2 x 2 ...

Question

$\int t a n^{3} 2 x s e c 2 x d x =$

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

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\frac{1}{6} s e c^{3} 2 x + \frac{1}{2} s e c 2 x + c

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\frac{1}{9} s e c^{2} 2 x - \frac{1}{3} s e c 2 x + c

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None of these

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Solution

The correct option is A $\frac{1}{6} s e c^{3} 2 x - \frac{1}{2} s e c 2 x + c$
$\int t a n^{3} 2 x s e c 2 x d x = \int (s e c^{2} 2 x - 1) s e c 2 x t a n 2 x d x$
$= \int (s e c^{3} 2 x t a n 2 x - s e c 2 x t a n 2 x) d x$
$= \int s e c^{3} 2 x t a n 2 x d x - \int s e c 2 x t a n 2 x d x$ ........ (i)
Now, we take $\int s e c^{3} 2 x t a n 2 x d x$
Put $s e c 2 x = t \Rightarrow s e c 2 x t a n 2 x = \frac{d t}{2}$ , then it reduces to
$\frac{1}{2} \int t^{2} d t = \frac{t^{3}}{6} = \frac{s e c^{3} 2 x}{6}$
From (i), $\int s e c^{3} 2 x t a n 2 x d x - \int s e c 2 x t a n 2 x d x$
$= \frac{s e c^{3} 2 x}{6} - \frac{s e c 2 x}{2} + c$
Trick: Let $s e c 2 x = t, t h e n s e c 2 x t a n 2 x d x = \frac{1}{2} d t ∴ \frac{1}{2} \int (t^{2} - 1) d t = \frac{1}{6} t^{3} - \frac{1}{2} t + c = \frac{1}{6} s e c^{3} 2 x - \frac{1}{2} s e c 2 x + c$

Suggest Corrections

∫tan32x sec 2x dx=

$\int t a n^{3} 2 x s e c 2 x d x =$