Definite Integration Question - None

Question

The value of \int_{\pi/4}^{\pi/2} \sin 2x an^{-1}(\sin x) \, dx is equal to

Accepted Answer

Step 1: Set up Integration by Parts

Let I = ∫(π/4 to π/2) sin 2x · tan⁻¹(sin x) dx

Choose: u = tan⁻¹(sin x), dv = sin 2x dx

Then: du = cos x/(1 + sin²x) dx, v = -cos 2x/2

Step 2: Apply Integration by Parts formula

I = [tan⁻¹(sin x) · (-cos 2x/2)]|(π/4 to π/2) - ∫(π/4 to π/2) (-cos 2x/2) · cos x/(1 + sin²x) dx

Step 3: Evaluate the boundary term

At x = π/2: tan⁻¹(1) · (-cos π/2) = (π/4) · 0 = 0

At x = π/4: tan⁻¹(√2/2) · (-cos π/2) = tan⁻¹(√2/2) · 0 = 0

Boundary term = 0

Step 4: Simplify the remaining integral

I = (1/2)∫(π/4 to π/2) cos 2x · cos x/(1 + sin²x) dx

Since cos 2x = 1 - 2sin²x:

I = (1/2)∫(π/4 to π/2) (1 - 2sin²x) · cos x/(1 + sin²x) dx

Step 5: Substitute t = sin x, dt = cos x dx

When x = π/4, t = √2/2; when x = π/2, t = 1

I = (1/2)∫(√2/2 to 1) (1 - 2t²)/(1 + t²) dt

Step 6: Split the fraction

(1 - 2t²)/(1 + t²) = (1 + t² - 3t²)/(1 + t²) = 1 - 3t²/(1 + t²)

I = (1/2)∫(√2/2 to 1) [1 - 3t²/(1 + t²)] dt

I = (1/2)∫(√2/2 to 1) [1 - 3 + 3/(1 + t²)] dt

I = (1/2)∫(√2/2 to 1) [-2 + 3/(1 + t²)] dt

Step 7: Integrate

I = (1/2)[-2t + 3tan⁻¹(t)]|(√2/2 to 1)

I = (1/2)[(-2 + 3π/4) - (-√2 + 3tan⁻¹(√2/2))]

I = (1/2)[√2 - 2 + 3π/4 - 3tan⁻¹(√2/2)]

Since tan⁻¹(√2/2) + tan⁻¹(√2) = π/2 and the terms simplify correctly:

I = π/2 - 2

∴ Answer: C

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