-1-2 tan xtan x- x1 - 2 d x-A- log - x x-tan x-cB- log - x x-tan x-cC- log- x- x-tan x-cD- log -cos x x-tan x-c

The correct option is B log |sec x(sec x + tan x)|+cI=∫{1+2 tan x(tan x+sec x)}^1/2 dx=∫ [1+2 tan^2 x+2 tan x sec x]^1/2 dx =∫ (sec^2 x+tan^2 x+2 sec x tan x ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

∫1+2 tan xtan...

Question

$\int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x =$

l o g | s e c x (s e c x - t a n x) | + c

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l o g | s e c x (s e c x + t a n x) | + c

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l o g ∣ ∣ \frac{s e c x}{s e c x + t a n x} ∣ ∣ + c

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l o g | c o s x (s e c x + t a n x) | + c

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Solution

The correct option is B $l o g | s e c x (s e c x + t a n x) | + c$
$I = \int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x = \int [1 + 2 t a n^{2} x + 2 t a n x s e c x]^{1 / 2} d x = \int (s e c^{2} x + t a n^{2} x + 2 s e c x t a n x)^{1 / 2} d x = \int (s e c x + t a n x) d x = l o g (s e c x + t a n x) + l o g s e c x + c = l o g [s e c x (s e c x + t a n x)] + c$

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Similar questions

Integration of sec x is:

\int {1 + 2 tan x (tan x + sec x)}^{\frac{1}{2}} d x

is equal to

\int \frac{(1 + \sqrt{t a n x}) (1 + t a n^{2} x)}{2 t a n x} d x

equal to

$\int \frac{1}{2 + 3 s i n x} d x$

The general solution of $\frac{d y}{d x} + y t a n x = s e c x$ is
(a) y secx=tanx+C
(b) y tanx=secx+C
(c) tanx=y tanx+C
(d) x secx=tany+C

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∫{1+2 tan x(tan x+sec x)}1/2 dx=

The correct option is B log |sec x(sec x+tan x)|+cI=∫{1+2 tan x(tan x+sec x)}1/2 dx=∫[1+2 tan2 x+2 tan x sec x]1/2 dx=∫(sec2 x+tan2 x+2 sec x tan x)1/2 dx=∫(sec x+tan x) dx=log(sec x+tan x)+log sec x+c=log[sec x(sec x+tan x)]+c

$\int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x =$