-tan x-dx -1-2 -2- -2 tan-1 -tan-x-1-2-tan-x- - log -tan-2 x - 2 tan-x-1-tan-x-2 tan-x-1-c-substitute tanx - t-21-2 -2- -2 tan-1 -tan-x-1-2-tan-x- - log -tan-x - -2 tan x-1-tan-x - -2 tan-x-1-c1-2 -2- -2 tan-1 -tan-x-1-2-tan-x- - log -tan x - -2 tan-x-1-tan-x - -2 tan-x-1-c1-5 -2- -2-tan-1 -tan-x-1-2- tan x - - log -tan-2 x - 2-tan-x-1-tan-x-1-tan x-1-

The correct option is B 1/2 √(2) [2 tan^ 1 (tan x 1/√(2 tan x)) + log [tan x √(2 tan x)+1/tan x + √(2 tan x)+1]]+c1/2 2 ∫√(tan x) dx = 1/2∫ (√(tanx) + √(cot ...

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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 2

∫√(tan x) dx ...

Question

$\int \sqrt{t a n x} d x =$

\frac{1}{2 \sqrt{2}} [2 t a n^{- 1} (\frac{t a n x - 1}{\sqrt{2 t a n x}}) + l o g [\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}]] + c

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\frac{1}{2 \sqrt{2}} [2 t a n^{- 1} (\frac{t a n x - 1}{\sqrt{2} t a n x}) + l o g [\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}]] + c

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\frac{1}{5 \sqrt{2}} [2 t a n^{- 1} (\frac{t a n x + 1}{\sqrt{2} t a n x}) + l o g [\frac{t a n^{2} x - 2 t a n x + 1}{t a n x - 1 t a n x + 1}]]

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\frac{1}{2 \sqrt{2}} [2 t a n^{- 1} (\frac{t a n x + 1}{\sqrt{2} t a n x}) + l o g [\frac{t a n^{2} x - 2 t a n x - 1}{t a n x + 2 t a n x + 1}]] + c, s u b s t i t u t e t a n x = t^{2}

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Solution

The correct option is A $\frac{1}{2 \sqrt{2}} [2 t a n^{- 1} (\frac{t a n x - 1}{\sqrt{2 t a n x}}) + l o g [\frac{t a n x - \sqrt{2 t a n x} + 1}{t a n x + \sqrt{2 t a n x} + 1}]] + c$
$\frac{1}{2} 2 \int \sqrt{t a n x} d x = \frac{1}{2} \int (\sqrt{t a n x} + \sqrt{c o t x}) d x + \frac{1}{2} \int (\sqrt{t a n x} - \sqrt{c o t x}) d x, s u b s t i t u t e t a n x = t^{2}$

Suggest Corrections

∫√tanx dx=

The correct option is A 12√2[2tan−1(tan x−1√2 tan x)+log[tan x−√2tanx+1tan x+√2tan x+1]]+c122∫√tanx dx=12∫(√tanx+√cotx)dx+12∫(√tanx−√cotx)dx, substitute tan x=t2

$\int \sqrt{t a n x} d x =$