If ∫ √(sec 2x-1)dx=α logₑ|cos 2x+β+√(cos 2x(1+cos 1/βx))|+ constant, then β-α is equal to ____________.

Solution

Maths sir · 4 steps

Multiply inside the root by 1+cos 2x; the numerator becomes sin 2x, and t = cos 2x finishes it.

1. Rewrite the root. sec 2x - 1 = (1-cos 2x)/(cos 2x). Multiply the top and the bottom by 1+cos 2x. The top becomes 1-cos² 2x = sin² 2x, so the integrand is (sin 2x)/√(cos² 2x + cos 2x).
2. Substitute. Put t = cos 2x, so dt = -2 sin 2x dx. The integral becomes -1/2∫ dt/√(t²+t). Complete the square: t² + t = (t+1/2)² - 1/4.

   -1/2∫ dt/√((t+1/2)²-1/4) = -1/2 ln|t+1/2+√(t²+t)| + C

3. Match the form. Put t = cos 2x back: √(t²+t) = √(cos 2x (1+cos 2x)). This matches the given form with α = -1/2 and β = 1/2. Check: 1/βx = 2x, which agrees with cos 2x inside the root.
4. So. β - α = 1/2 - (-1/2) = 1.

The common slip: 0. Dropping the minus sign in dt = -2 sin 2x dx gives α = 1/2, so β - α = 0 instead of 1.

The idea to keep. For √((1-cos θ)/(cos θ)), multiply inside the root by 1+cos θ. The top becomes sin θ, the dt for t=cos θ, and a standard root form is left.

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