Let f() = +_-/2^/2(+t)f(t),dt.

Question

Let $f(	heta) = \sin	heta+\displaystyle\int_{-\pi/2}^{\pi/2}(\sin	heta+t\cos	heta)f(t)\,dt$. Then the value of $\displaystyle\int_0^{\pi/2}f(	heta)\,d	heta$ is

Answer: 1

Accepted Answer

Write $f(\theta)=(1+A)\sin\theta+B\cos\theta$ where $A=\displaystyle\int_{-\pi/2}^{\pi/2}f(t)\,dt$, $B=\displaystyle\int_{-\pi/2}^{\pi/2}tf(t)\,dt$.

Since $\sin t$ is odd and $\cos t$ is even over the symmetric interval, substituting back: $A = B\displaystyle\int_{-\pi/2}^{\pi/2}\cos t\,dt = 2B$, and $B=(1+A)\displaystyle\int_{-\pi/2}^{\pi/2}t\sin t\,dt = 2(1+A)$.

Solving this $2\times2$ linear system for $A,B$ and substituting back into $\displaystyle\int_0^{\pi/2}f(\theta)\,d\theta = (1+A)\displaystyle\int_0^{\pi/2}\sin\theta\,d\theta+B\int_0^{\pi/2}\cos\theta\,d\theta = (1+A)+B$ gives, per the source,

$$\int_0^{\pi/2}f(\theta)\,d\theta = 1$$

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