Integrate the following functions w-r-t- x- -1-cos x-a cos x-b d x

∫1/cos(x+a) cos(x+b)dx On multiplying and dividing by sin(a b), we get I=1/sin(a b)∫sin(a b)/cos(x+a) cos(x+b)dx ⇒ I=1/sin(a b)∫sin[(x+a) (x+b)]/cos(x+a) cos ...

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Standard XII

Mathematics

Standard Formulae - 1

Integrate the...

Question

Integrate the following functions w.r.t. x.

$\int \frac{1}{c o s (x + a) c o s (x + b)} d x .$

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Solution

$\int \frac{1}{c o s (x + a) c o s (x + b)} d x$

On multiplying and dividing by sin(a - b), we get

$I = \frac{1}{s i n (a - b)} \int \frac{s i n (a - b)}{c o s (x + a) c o s (x + b)} d x \Rightarrow I = \frac{1}{s i n (a - b)} \int \frac{s i n [(x + a) - (x + b)]}{c o s (x + a) c o s (x + b)} = \frac{1}{s i n (a - b)} \int \frac{s i n (x + a) c o s (x + b) - c o s (x + a) s i n (x + b)}{c o s (x + a) c o s (x + b)} d x [∵ s i n (A - B) = s i n A c o s B - s i n B c o s A] = \frac{1}{s i n (a - b)} \int [\frac{s i n (x + a)}{c o s (x + a)} - \frac{s i n (x + b)}{c o s (x + b)}] d x = \frac{1}{s i n (a - b)} \int [t a n (x + a) - t a n (x + b)] d x = \frac{1}{s i n (a - b)} [- l o g | c o s (x + a) | + l o g | c o s (x + b) |] + C = \frac{1}{s i n (a - b)} l o g ∣ ∣ \frac{c o s (x + b)}{c o s (x + a)} ∣ ∣ + C (∵ l o g m - l o g n = l o g \frac{m}{n})$

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Similar questions

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Integrate the following functions w.r.t. x. ∫1cos(x+a) cos(x+b)dx.

Integrate the following functions w.r.t. x.

$\int \frac{1}{c o s (x + a) c o s (x + b)} d x .$