Primitive of (cos⁡x−sin⁡x)15−sin⁡2x is equal to

I=∫cosx−sinxdx15−sin2x Put sin⁡x+cos⁡x=t(cos⁡x−sin⁡x)dx=dtsin⁡x+cos⁡x=t1+2sin⁡xcos⁡x=t2sin⁡2x=t2−1I=∫dt15−t2−1I=∫dt15−t2−1=sin−1⁡t4+c=sin−1⁡sin⁡x+cos⁡x4+c

Primitive of (cos⁡x−sin⁡x)15−sin⁡2x is equal to

I=∫cosx−sinxdx15−sin2x Put sin⁡x+cos⁡x=t(cos⁡x−sin⁡x)dx=dtsin⁡x+cos⁡x=t1+2sin⁡xcos⁡x=t2sin⁡2x=t2−1I=∫dt15−t2−1I=∫dt15−t2−1=sin−1⁡t4+c=sin−1⁡sin⁡x+cos⁡x4+c

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$Primitive of \frac{(\cos x - \sin x)}{\sqrt{15 - \sin 2 x}} is equal to$

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$\sin^{- 1} (\frac{\sin \frac{x}{2} - \cos \frac{x}{2}}{4}) + c$

$\sin^{- 1} (\frac{\sin x + \cos x}{4}) + c$

$\sin^{- 1} (\frac{\sin x - \cos x}{4}) + c$

$\sin^{- 1} (\frac{\cos x - \sin x}{4}) + c$

answer is B.

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Detailed Solution

$I = \int \frac{(\cos x - \sin x) d x}{\sqrt{15 - \sin 2 x}}$

$Put \sin x + \cos x = t$

$\begin{array}{l} (\cos x - \sin x) d x = d t \\ \sin x + \cos x = t \\ 1 + 2 \sin x \cos x = t^{2} \\ \sin 2 x = t^{2} - 1 \\ I = \int \frac{d t}{\sqrt{15 - (t^{2} - 1)}} \\ I = \int \frac{d t}{\sqrt{15 - (t^{2} - 1)}} = \sin^{- 1} (\frac{t}{4}) + c \\ = \sin^{- 1} (\frac{\sin x + \cos x}{4}) + c \end{array}$