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Maths-

General

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Question

If $not stretchy integral fraction numerator sin invisible function application x over denominator sin invisible function application left parenthesis x minus alpha right parenthesis end fraction d x equals$ Ax + B log sin(x – $alpha$ ) + c, then value of (A, B) is –

(– sin $α$ , cos $α$ )
(cos $α$ , sin $α$ )
(sin $α$ , cos $α$ )
(– cos $α$ , sin $α$ )

hint Hint:

solve the problem by replacing x with $y space plus alpha$ then evaluate and after solving re substitute the value of
y = x - $alpha$ then compare given and obtained equations to find the value of A and B.

The correct answer is: (cos $alpha$ , sin $alpha$ )

Given , $not stretchy integral fraction numerator sin invisible function application x over denominator sin invisible function application left parenthesis x minus alpha right parenthesis end fraction d x equals$ Ax + B log sin(x – $alpha$ ) + c,
substitute x = y + $alpha$ then dx = dy
$not stretchy integral fraction numerator sin invisible function application x over denominator sin invisible function application left parenthesis x minus alpha right parenthesis end fraction d x equals$ $integral fraction numerator sin invisible function application left parenthesis y plus alpha right parenthesis over denominator sin invisible function application left parenthesis y plus alpha minus alpha right parenthesis end fraction d x equals space integral fraction numerator sin invisible function application left parenthesis y plus alpha right parenthesis over denominator sin invisible function application left parenthesis y right parenthesis end fraction d y space$
Using sin (a + b) = sin a cos b + sin b cos a
$integral fraction numerator sin space y space cos space alpha space plus space sin space alpha space cos space y space over denominator sin space y end fraction space d y space equals space cos space alpha integral 1. space d y space plus sin space alpha integral space c o t space y space d y w e space k n o w comma integral space c o t space y space d x space equals space ln open vertical bar sin space y space close vertical bar plus space c t h e n comma space space cos space alpha integral 1. space d y space plus sin space alpha integral space c o t space y space d y space equals space y space cos space alpha space plus space sin space alpha space ln open vertical bar sin space y close vertical bar plus space c$
As x = y + $alpha$ the y = x - $alpha$ then,
$not stretchy integral fraction numerator sin invisible function application x over denominator sin invisible function application left parenthesis x minus alpha right parenthesis end fraction d x equals$ $left parenthesis x minus alpha right parenthesis space cos space alpha space plus space sin space alpha space ln open vertical bar sin space left parenthesis x minus alpha right parenthesis close vertical bar plus space k space$
$equals space x space cos space alpha space plus space sin space alpha space ln open vertical bar sin space left parenthesis x minus alpha right parenthesis close vertical bar space plus space c space space w h e r e space c space i s space a space n e w space a r b i t a r y space c o n s tan t$
Upon comparing we get A = cos $alpha$ and B = sin $alpha$
Therefore, value of (A, B) = ( cos $alpha$ , sin $alpha$ )

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