-d x-cos x-a cos x-b-A- coseca-b logsin x-a-sin x-b-cB- coseca-b logcos x-a-cos x-b-cC- coseca-b logsin x-b-sin x-a-cD- coseca-b logcos x-b-cos x-a-c

The correct option is B cosec(a b) log cos(x a)/cos(x b)+c∫dx/cos(x a)cos(x b) =1/sin(a b)∫sin{(x b) (x a)}/cos(x a).cos(x b)dx =1/sin(a b)∫{sin(x b)/cos(x ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Common Angles

∫d x/cos x-a ...

Question

$\int \frac{d x}{c o s (x - a) c o s (x - b)} =$

c o s e c (a - b) l o g \frac{s i n (x - a)}{s i n (x - b)} + c

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c o s e c (a - b) l o g \frac{c o s (x - a)}{c o s (x - b)} + c

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c o s e c (a - b) l o g \frac{s i n (x - b)}{s i n (x - a)} + c

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c o s e c (a - b) l o g \frac{c o s (x - b)}{c o s (x - a)} + c

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Solution

The correct option is B $c o s e c (a - b) l o g \frac{c o s (x - a)}{c o s (x - b)} + c$
$\int \frac{d x}{c o s (x - a) c o s (x - b)} = \frac{1}{s i n (a - b)} \int \frac{s i n {(x - b) - (x - a)}}{c o s (x - a) . c o s (x - b)} d x = \frac{1}{s i n (a - b)} \int {\frac{s i n (x - b)}{c o s (x - b)} - \frac{s i n (x - a)}{c o s (x - a)}} d x = c o s e c (a - b) l o g \frac{c o s (x - a)}{c o s (x - b)} + c$

Suggest Corrections

∫dxcos(x−a)cos(x−b)=

The correct option is B cosec(a−b) logcos(x−a)cos(x−b)+c∫dxcos(x−a)cos(x−b)=1sin(a−b)∫sin{(x−b)−(x−a)}cos(x−a).cos(x−b)dx=1sin(a−b)∫{sin(x−b)cos(x−b)−sin(x−a)cos(x−a)}dx=cosec(a−b) logcos(x−a)cos(x−b)+c

$\int \frac{d x}{c o s (x - a) c o s (x - b)} =$