The correct option is B 1/2(sec x+tan x)^2+cPut t=sec x+tan x ⇒dt/t = sec x dx ∫sec x/(sec x+tan x)^2 dx =∫1/t^2dt/t = ∫ t^ 3dt=t^ 2/ 2+c = 1/2(sec x+tan x) ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫ x/ x+tan x2...

Question

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

- \frac{1}{2 (s e c x + t a n x)^{2}} + c

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\frac{1}{(s e c x + t a n x)^{2}} + c

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\frac{1}{2 (s e c x - t a n x)^{2}} + c

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None

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Solution

The correct option is A $- \frac{1}{2 (s e c x + t a n x)^{2}} + c$
Put $t = s e c x + t a n x \Rightarrow \frac{d t}{t} = s e c x d x$
$\int \frac{s e c x}{(s e c x + t a n x)^{2}} d x = \int \frac{1}{t^{2}} \frac{d t}{t} = \int t^{- 3} d t = \frac{t^{- 2}}{- 2} + c = \frac{- 1}{2 (s e c x + t a n x)^{2}} + c$

Suggest Corrections

∫sec x(sec x+tan x)2 dx=

The correct option is A −12(sec x+tan x)2+cPut t=sec x+tan x⇒dtt=sec x dx ∫sec x(sec x+tan x)2 dx=∫1t2dtt=∫t−3dt=t−2−2+c=−12(sec x+tan x)2+c

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