Question

Let f be integrable over [0, a] for any real a. If we define I₁ = $\int_{0}^{\pi/2}{\cos\theta}$f(sin q + cos²q) dq and

I₂ =$\int_{0}^{\pi/2}{\sin 2\theta}$f(sin q + cos²q)dq, then

• A. I₁ = I₂
• B. I₁ = –I₂
• C. I₁ = 2I₂
• D. I₁ = –2I₂

Answer 1

Answer: (A)

I₁ = I₂

Answer 2

I₁ = $\int_{0}^{\frac{\pi}{2}}{\cos\theta}$f(sinq + cos2q) dq

I₂ = $\int_{0}^{\frac{\pi}{2}}{2\sin\theta\cos\theta}$f(sinq + cos²q) dq

Let sinq = t, then I₁ = $\int_{0}^{1}{f(t + 1–t^{2})}$dt.

and I₂ = 2$\int_{0}^{1}{tf(t + 1–t^{2})dt}$

replace t by 1 – t then I₁ = I₂