Mathematics

# $\int \sqrt { \sec x - 1 } d x$ is equal to

##### SOLUTION
$I=\displaystyle \int (\sqrt{\sec x - 1}) dx$

$I=\displaystyle \int \sqrt{\dfrac{1}{\cos x} -1} = \int \sqrt{\dfrac{1 -\cos x}{\cos x}} dx$

$1 - cos 2x = 2 \sin^2 (x)$

$\cos 2x = 2 \cos^2 x - 1$

$I=\displaystyle \int \sqrt{\dfrac{2 \sin^2 (x/2)}{2 \cos^2 (x/2) - 1}} dx$

$u = \cos x/2 \,\,\, du = -\sin (x/2) \times \dfrac{1}{2} dx$

$I=\displaystyle \int \sqrt{\dfrac{2 \times (\sin^2 (x /2))}{2u^2 - 1}} \times \dfrac{du}{\dfrac{-1}{2} \times \sin (x/2)}$

$I=-2\sqrt{2} \displaystyle \int \dfrac{1}{\sqrt{2u^2 - 1}} du$

$I=\dfrac{-2 \sqrt{2}}{\sqrt{2}} \displaystyle \int \dfrac{1}{\sqrt{u^2 - \dfrac{1}{2}}} du$

$I=\displaystyle \int \dfrac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + c$

$I=-2 \log |u + \sqrt{u^2 - 1/2} | + c$

$I=-2 \log |\cos x/2 + \sqrt{\cos^2 \dfrac{x}{2} - \dfrac{1}{2}} | + c$.

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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