int_0^{frac{pi}{2}} frac{300 sin x+100 cos x} | vedclass

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(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{300 \sin x + 100 \cos x}{\sin x + \cos x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, w

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(A) Let $I = \int_0^{\frac{\pi}{2}} \frac{300 \sin x + 100 \cos x}{\sin x + \cos x} dx$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$, we get: $I = \int_0^{\frac{\pi}{2}} \frac{300 \sin(\frac{\pi}{2}-x) + 100 \cos(\frac{\pi}{2}-x)}{\sin(\frac{\pi}{2}-x) + \cos(\frac{\pi}{2}-x)} dx = \int_0^{\frac{\pi}{2}} \frac{300 \cos x + 100 \sin x}{\cos x + \sin x} dx$. Adding the two expressions for $I$: $2I = \int_0^{\frac{\pi}{2}} \frac{(300 \sin x + 100 \cos x) + (300 \cos x + 100 \sin x)}{\sin x + \cos x} dx$. $2I = \int_0^{\frac{\pi}{2}} \frac{400 \sin x + 400 \cos x}{\sin x + \cos x} dx$. $2I = \int_0^{\frac{\pi}{2}} 400 dx$. $2I = 400 [x]_0^{\frac{\pi}{2}} = 400 	imes \frac{\pi}{2} = 200 \pi$. Therefore, $I = 100 \pi$.

$\int_0^{\frac{\pi}{2}} \frac{300 \sin x+100 \cos x}{\sin x+\cos x} \,dx = \ldots$ (in $\pi$)

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