Let f(theta) = sin theta + int_{-pi / 2}^{pi | vedclass

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(D) Given $f(	heta) = \sin 	heta + \int_{-\pi / 2}^{\pi / 2} (\sin 	heta + t \cos 	heta) f(t) dt$.We can rewrite this as $f(	heta) = \sin 	heta

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(D) Given $f(	heta) = \sin 	heta + \int_{-\pi / 2}^{\pi / 2} (\sin 	heta + t \cos 	heta) f(t) dt$.We can rewrite this as $f(	heta) = \sin 	heta + \sin 	heta \int_{-\pi / 2}^{\pi / 2} f(t) dt + \cos 	heta \int_{-\pi / 2}^{\pi / 2} t f(t) dt$.Let $A = \int_{-\pi / 2}^{\pi / 2} f(t) dt$ and $B = \int_{-\pi / 2}^{\pi / 2} t f(t) dt$.Then $f(	heta) = (A + 1) \sin 	heta + B \cos 	heta$.Now,calculate $A = \int_{-\pi / 2}^{\pi / 2} ((A + 1) \sin t + B \cos t) dt$. Since $\sin t$ is an odd function,$\int_{-\pi / 2}^{\pi / 2} \sin t dt = 0$. Thus,$A = B \int_{-\pi / 2}^{\pi / 2} \cos t dt = B [\sin t]_{-\pi / 2}^{\pi / 2} = 2B$.Next,calculate $B = \int_{-\pi / 2}^{\pi / 2} t ((A + 1) \sin t + B \cos t) dt$. Since $t \cos t$ is an odd function,$\int_{-\pi / 2}^{\pi / 2} t \cos t dt = 0$. Thus,$B = (A + 1) \int_{-\pi / 2}^{\pi / 2} t \sin t dt = (A + 1) \cdot 2 \int_{0}^{\pi / 2} t \sin t dt$.Using integration by parts,$\int_{0}^{\pi / 2} t \sin t dt = [-t \cos t]_{0}^{\pi / 2} + \int_{0}^{\pi / 2} \cos t dt = 0 + [\sin t]_{0}^{\pi / 2} = 1$.So,$B = 2(A + 1)$.Substituting $A = 2B$ into $B = 2(2B + 1)$,we get $B = 4B + 2$,which implies $3B = -2$,so $B = -2/3$ and $A = -4/3$.Then $f(	heta) = (-4/3 + 1) \sin 	heta - (2/3) \cos 	heta = -1/3 \sin 	heta - 2/3 \cos 	heta$.Finally,$\left| \int_{0}^{\pi / 2} f(	heta) d	heta ight| = \left| \int_{0}^{\pi / 2} (-1/3 \sin 	heta - 2/3 \cos 	heta) d	heta ight| = \left| [-1/3(-\cos 	heta) - 2/3 \sin 	heta]_{0}^{\pi / 2} ight| = \left| [1/3 \cos 	heta - 2/3 \sin 	heta]_{0}^{\pi / 2} ight| = \left| (0 - 2/3) - (1/3 - 0) ight| = \left| -2/3 - 1/3 ight| = |-1| = 1$.

Let $f(\theta) = \sin \theta + \int_{-\pi / 2}^{\pi / 2} (\sin \theta + t \cos \theta) f(t) dt$. Then the value of $\left| \int_{0}^{\pi / 2} f(\theta) d\theta \right|$ is

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