If x/(x-α)dx=Ax+B(x-α)+C, then value of (A, B) is - Mathematics MCQ AIEEE 2004 2004 | ExamTest.live

mediumMCQAIEEE 20042004Mathematics

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If \int \frac{s i n x}{s i n ( x - α )} d x = A x + B lo g sin (x - α) + C, then value of (A, B) is

Solution & Step-by-step Explanation

Let t = x - α \Rightarrow x = t + α and d x = d t . The integral becomes: \int \frac{s i n ( t + α )}{s i n t} d t = \int \frac{s i n t c o s α + c o s t s i n α}{s i n t} d t . = \int (cos α + sin α cot t) d t . = cos α \int d t + sin α \int cot t d t . = t cos α + sin α ln ∣ sin t ∣ + [c i t e_{s} t a r t] C . Substituting t = x - α : = (x - α) cos α + sin α ln ∣ sin (x - α) ∣ + [c i t e_{s} t a r t] C . = x cos α + sin α ln ∣ sin (x - α) ∣ + (C - α cos α) [c i t e_{s} t a r t] . Comparing with the form A x + B ln sin (x - α) + C^{'} : A = cos α and B = sin α . Thus, (A, B) = (cos α, sin α) .

Try it yourself before checking the explanation above.

If \int \frac{s i n x}{s i n ( x - α )} d x = A x + B lo g sin (x - α) + C, then value of (A, B) is

(sin α, cos α)

(cos α, sin α)

(- sin α, cos α)

(- cos α, sin α)

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