If $\int \frac{c o s x - s i n x}{\sqrt[]{8 - s i n 2 x}} d x = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c,$ where c is a constant of integration, then the ordered pair (a, b) is equal to:

Option 1 -

$(- 1, 3)$

Option 2 -

$(1, - 3)$

Option 3 -

$(3, 1)$

Option 4 -

$(1, 3)$

2 Views | Posted a month ago

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1 Answer

Answered by

alok kumar singh | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

$\int \frac{(c o s x - s i n x) d x}{\sqrt[]{8 - s i n 2 x}} = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c$

$\int \frac{(c o s x - s i n x) d x}{\sqrt[]{9 - {(s i n x + c o s x)}^{2}}} =$

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$= \int \frac{d t}{\sqrt[]{3^{2} - t^{2}}} = s i n^{- 1} \frac{t}{3} + c = s i n^{- 1} (\frac{s i n x + c o s x}{3}) + c$

= a = 1, b = 3

$∴ (a, b) \equiv (1, 3)$

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If $\int \frac{c o s x - s i n x}{\sqrt[]{8 - s i n 2 x}} d x = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c,$ where c is a constant of integration, then the ordered pair (a, b) is equal to:

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If ∫ c o s x − s i n x 8 − s i n 2 x d x = a s i n − 1 ( s i n x + c o s x b ) + c , where c is a constant of integration, then the ordered pair (a, b) is equal to:

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If $\int \frac{c o s x - s i n x}{\sqrt[]{8 - s i n 2 x}} d x = a s i n^{- 1} (\frac{s i n x + c o s x}{b}) + c,$ where c is a constant of integration, then the ordered pair (a, b) is equal to: