If ∫cosx-sinx8-sin2xdx=asin-1(sinx+cosxb)+c. where c is a constant of integration, then the ordered pair (a, b) is equal to ______. - Mathematics (JEE Main)

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If `int(cosx - sinx)/sqrt(8 - sin2x)dx = asin^-1((sinx + cosx)/b) + c`. where c is a constant of integration, then the ordered pair (a, b) is equal to ______.

Options

  • (1, –3)

  • (1, 3)

  • (–1, 3)

  • (3, 1)

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Solution

If `int(cosx - sinx)/sqrt(8 - sin2x)dx = asin^-1((sinx + cosx)/b) + c`. where c is a constant of integration, then the ordered pair (a, b) is equal to (1, 3).

Explanation:

Given: I = `int(cosx - sinx)/sqrt(8 - sin2x)dx = asin^-1((sinx + cosx)/b) + c`

Let, sinx + cosx = t  ....(i)

Squaring both sides, we get

(sinx + cosx)2 = t

⇒ sin2x + cos2x + 2sinxcosx = t2

⇒ 1 + 2sinxcosx = t2  ...(∵ sin2x + cos2x = 1)

⇒ sin2x = t2 = 1   ...(∵ 2sinxcosx = sin2x)

Differentiate equation (i) both sides w.r.t.x

(cosx – sinx)dx = dt  ....(ii)

Putting value we get,

I = `int(dt)/sqrt(9 - t^2) = sin^-1(t/3) + c`  ...`(∵ int (dx)/sqrt(a^2 - x^2) = sin^-1(x/a) + c)`

= `sin^-1((sinx + cosx)/3) + c`

Compare with `asin^-1((sinx + cosx)/b) + c`

∴ a = 1 and b = 3

Ordered pair (a, b) = (1, 3).

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