First slide

Methods of integration

Question

If $\int \sqrt{1 + \sec x} d x = K \sin^{- 1} (f (x)) + C$ then

Easy

A

$f (x) = \sqrt{2} \sin (x / 2), k = 2$
B

$f (x) = \sqrt{2} \cos (x / 2), K = 2$
C

$f (x) = \sqrt{2} \tan (x / 2), K = 2$
D

$f (x) = \sqrt{2} \sin (x / 2), K = \sqrt{2}$

Solution

$\sqrt{1 + \sec x} = \sqrt{\frac{1 + \cos x}{\cos x}} = \frac{\sqrt{2} \cos x / 2}{\sqrt{1 - 2 \sin^{2} x / 2}}$ so
putting $\sin x / 2 = t$ we have
$\begin{array}{l} \int \sqrt{1 + \sec x} d x = \int \frac{\sqrt{2} \cos x / 2}{\sqrt{1 - 2 \sin^{2} x / 2}} d x \\ = 2 \sin^{- 1} \sqrt{2} t + C (t = \sin x / 2) \\ = 2 \sin^{- 1} (\sqrt{2} \sin x / 2) + C \end{array}$
so $f (x) = \sqrt{2} \sin x / 2 and K = 2$

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