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Q22  Find the integrals of the functions \frac{1}{\cos ( x-a ) \cos ( x-b )}

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Using the trigonometric identities following integrals can be simplified as follows

\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}]

                                               =\frac{1}{\sin(a-b)}[\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}]

                                               =\frac{1}{\sin(a-b)}[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}]

=\frac{tan(x-b)-\tan (x-a)}{\sin(a-b)}

=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx
\\=\frac{1}{\sin(a-b)}[-\log\left | \cos(x-b) \right |+\log\left | \cos(x-a) \right |]\\ =\frac{1}{\sin(a-b)}(\log\left | \frac{\cos(x-a)}{\cos(x-b)} \right |)

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manish

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