Question

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 ______.

पर्याय

• (1, –3)

• (1, 3)

• (–1, 3)

• (3, 1)

Answer 1

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).