Question

If ${\displaystyle \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c}$. 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)

Answer 1

If ${\displaystyle \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c}$. where c is a constant of integration, then the ordered pair (a, b) is equal to (1, 3).

Explanation:

Given: I = ${\displaystyle \int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c}$

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 = ${\displaystyle \int \frac{dt}{\sqrt{9-t^2}}={\sin }^{-1}\left(\frac{t}{3}\right)+c}$  ...${\displaystyle (\because \int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\left(\frac{x}{a}\right)+c)}$

= ${\displaystyle {\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c}$

Compare with ${\displaystyle a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c}$

∴ a = 1 and b = 3

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