-dx-sin x-sinx- is equal to

∫dx/sin x. sin(x+α) =1/sin α∫sin((x+α) x)/sin x. sin(x+α)dx =cosec α∫(sin(x+α)cos x cos(x+α)sin x)/sin x. sin(x+α)dx =cosec α∫ [cot x cot(x+α)]dx =cosec ...

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Byju's Answer

Standard XIII

Mathematics

Standard Formulae - 2

∫dx/sin x.sin...

Question

$\int \frac{d x}{s i n x . s i n (x + α)}$ is equal to

$c o s e c α l o g ∣ ∣ \frac{s i n x}{s i n (x + α)} ∣ ∣ + C$

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$c o s e c α l o g ∣ ∣ \frac{s i n (x + α)}{s i n x} ∣ ∣ + C$

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$c o s e c α l o g ∣ ∣ \frac{s e c (x + α)}{s e c x} ∣ ∣ + C$

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$c o s e c α l o g ∣ ∣ \frac{s e c x}{s e c (x + α)} ∣ ∣ + C$

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Solution

The correct option is A
$c o s e c α l o g ∣ ∣ \frac{s i n x}{s i n (x + α)} ∣ ∣ + C$

$\int \frac{d x}{s i n x . s i n (x + α)} = \frac{1}{s i n α} \int \frac{s i n ((x + α) - x)}{s i n x . s i n (x + α)} d x = c o s e c α \int \frac{(s i n (x + α) c o s x - c o s (x + α) s i n x)}{s i n x . s i n (x + α)} d x = c o s e c α \int [c o t x - c o t (x + α)] d x = c o s e c α [l o g s i n x - l o g s i n (x + α)] + C = c o s e c α l o g ∣ ∣ \frac{s i n x}{s i n (x + α)} ∣ ∣ + C$

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∫dxsin x.sin(x+α) is equal to

$\int \frac{d x}{s i n x . s i n (x + α)}$ is equal to