What is the integral of $\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$ ?

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What is the integral of $\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$ ?

$\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$

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Answer 1 · 2018-05-12T05:02:56

The answer of this question= ${sin}^{- 1} (\frac{tan x}{\sqrt{3}})$ $sin^(-1)(tanx/sqrt3)$

Explanation:

For this take tanx=t
Then ${sec}^{2} x d x = d t$ $sec^2x dx=dt$

Also ${sec}^{2} x = 1 + {tan}^{2} x$ $sec^2x= 1+tan^2x$
Putting these value in original equation we get
$\int \frac{d t}{\sqrt{3 - t^{2}}}$ $intdt/(sqrt(3-t^2))$

$= {sin}^{- 1} (\frac{t}{\sqrt{3}})$ $=sin^(-1)(t/sqrt3)$

= ${sin}^{- 1} (\frac{tan x}{\sqrt{3}})$ $sin^(-1)(tanx/sqrt3)$

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What is the integral of $\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$ ?

$\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$

1 Answer

Explanation:

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What is the integral of ∫sec2x√4−sec2xdxint(sec^2x)/sqrt(4-sec^2x)dx ?

∫sec2x√4−sec2xdxint(sec^2x)/sqrt(4-sec^2x)dx

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Explanation:

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What is the integral of $\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$ ?

$\int \frac{{sec}^{2} x}{\sqrt{4 - {sec}^{2} x}} d x$ $int(sec^2x)/sqrt(4-sec^2x)dx$