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Standard Formulae - 1

Find the inte...

Question

Find the integrals of the functions.
$\int \frac{1}{c o s (x - a) c o s (x - b)} d x .$

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Solution

$\int \frac{1}{c o s (x - a) c o s (x - b)} d x = \frac{1}{s i n (b - a)} \int \frac{s i n (x - a) - (x - b)}{c o s (x -) c o s (x - b)} d x$
Multiply by sin (b-a)in numerator and denominator both and sin (b-a)=sin [(x-a)-(x-b)].
$= \frac{1}{s i n (b - a)} \int \frac{s i n (x - a) c o s (x - b) - c o s (x - a) s i n (x - b)}{c o s (x - a) c o s (x - b)} d x [∴ s i n (C - D) = s i n C c o s D - c o s C s i n D] = \frac{1}{s i n (b - a)} \int [\frac{s i n (x - a) c o s (x - b)}{c o s (x - a) c o s (x - b)} - \frac{c o s (x - a) s i n (x - b)}{c o s (x - a) c o s (x - b)}] d x = \frac{1}{s i n (b - a)} \int (t a n (x - a) - t a n (x - b)) d x = \frac{1}{s i n (b - a)} . {- l o g | c o s (x - a) | + l o g | c o s (x - b) |} + C . | ∴ \int t a n x d x = - l o g | c o s x | | = \frac{1}{s i n (b - a)} l o g ∣ ∣ \frac{c o s (x - b)}{c o s (x - a)} ∣ ∣ + C$

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