I-1sin x-a sin x-b dx-

The correct options are A 1sin ( a b )logsin ( x a )sin ( x b ). B 1sin ( b a )logsin ( x b )sin ( x a ). I=∫1/sin ( x a )sin ( x b )dx. ...

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Byju's Answer

Standard XII

Mathematics

Theorems for Continuity

I=∫1sin x-a s...

Question

$I = \int \frac{1}{sin (x - a) sin (x - b)} d x$ =.......................................

\frac{1}{sin (a - b)} log \frac{sin (x - a)}{sin (x - b)} .

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\frac{1}{sin (b - a)} log \frac{sin (x - b)}{sin (x - a)} .

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\frac{1}{cos (a - b)} log \frac{sin (x - b)}{sin (x - a)} .

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\frac{1}{cos (a - b)} log \frac{sin (x - a)}{sin (x - b)} .

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Solution

The correct options are
A $\frac{1}{sin (a - b)} log \frac{sin (x - a)}{sin (x - b)} .$
B $\frac{1}{sin (b - a)} log \frac{sin (x - b)}{sin (x - a)} .$
$I = \int \frac{1}{sin (x - a) sin (x - b)} d x .$
Substitute, $a - b = (x - b) - (x - a) \Rightarrow sin (a - b) = sin (x - b) cos (x - a) - cos (x - b) sin (x - a)$ ... (i)
$∴ I = \frac{1}{sin (a - b)} \int \frac{sin (x - b) cos (x - a) - cos (x - b) sin (x - a)}{sin (x - a) sin (x - b)} d x$ by (i) $= \frac{1}{sin (a - b)} \int [cot (x - a) - cot (x - b)] d x = \frac{1}{sin (a - b)} [log sin (x - a) - log sin (x - b)]$ $= \frac{1}{sin (a - b)} log \frac{sin (x - a)}{sin (x - b)} .$

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I=∫1sin(x−a)sin(x−b)dx=.......................................

$I = \int \frac{1}{sin (x - a) sin (x - b)} d x$ =.......................................