Question

cos x - sinx
√8 - sin 2x
dx =​

Answer 1

Question :-

$\rm{ \int \frac{ \cos x - \sin x }{ \sqrt{8 - \sin 2x } } dx = }$

Solution :

Putting cos x + sin x = t in the integral

$\sf \implies{( \cos x - \sin x)dx = dt }$

$\sf \implies{( \cos x - \sin x) {}^{2} = t {}^{2} }$

$\sf\implies{ \cos {}^{2}x + \sin {}^{2} x + 2 \sin x \cos x = t {}^{2} }$

$\sf\implies{1 + \sin 2x = t {}^{2} }$

$\rm{ \int \frac{dx}{ \sqrt{9 - (1 + \sin2x) } } } \\ \\ \\ \rm{ \int \frac{dt}{ \sqrt{9 - t {}^{2} } } } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \implies{ \sin {}^{ - 1} \frac{t}{3} + c }$

Answer :

$\small\rm{\int \frac{ \cos x - \sin x }{ \sqrt{8 - \sin2x } }dx = \sin {}^{ - 1} \frac{1 + \sin2x }{3} c }$

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