If ∫ (cos x-sin x)/√(8-sin 2x)dx=asin⁻¹((sin x+cos x)/b)+c, where c is a constant of integration, then the ordered pair (a, b) is equal to :

Solution

Maths sir · 3 steps

The numerator is the derivative of sin x + cos x, so put that equal to t.

1. Choose the substitution. Let t = sin x + cos x. Then dt = (cos x - sin x) dx, which is exactly the numerator.
2. Rewrite the root. Squaring gives t² = sin² x + cos² x + 2 sin x cos x = 1 + sin 2x. So sin 2x = t² - 1 and 8 - sin 2x = 9 - t².

   ∫ dt/√(3² - t²) = sin⁻¹(t/3) + c = sin⁻¹((sin x + cos x)/3) + c

3. So. Comparing with a sin⁻¹((sin x + cos x)/b) + c gives a = 1 and b = 3, so the ordered pair is (1, 3), option (b).

The idea to keep. When you see sin x ± cos x next to sin 2x, substitute t = sin x ± cos x. Its square is 1 ± sin 2x, and that turns the integral into a standard form.

Answer from the NTA final answer key. Solution by PrepWiser.