int_0^{pi / 2} frac{200 sin x+100 cos x}{sin | vedclass

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Question

(C) Let $I = \int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$. We can rewrite the numerator as $100(\sin x + \cos x) + 100 \sin x$. S

Vedclass · Accepted Answer

$75$

Vedclass · Answer

(C) Let $I = \int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$. We can rewrite the numerator as $100(\sin x + \cos x) + 100 \sin x$. So, $I = \int_0^{\pi / 2} \frac{100(\sin x + \cos x) + 100 \sin x}{\sin x + \cos x} d x = 100 \int_0^{\pi / 2} 1 d x + 100 \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$. Let $I_1 = \int_0^{\pi / 2} \frac{\sin x}{\sin x + \cos x} d x$ ... $(i)$. Using the property $\int_0^a f(x) d x = \int_0^a f(a-x) d x$, we get: $I_1 = \int_0^{\pi / 2} \frac{\sin(\pi/2 - x)}{\sin(\pi/2 - x) + \cos(\pi/2 - x)} d x = \int_0^{\pi / 2} \frac{\cos x}{\cos x + \sin x} d x$ ... (ii). Adding $(i)$ and (ii): $2I_1 = \int_0^{\pi / 2} \frac{\sin x + \cos x}{\sin x + \cos x} d x = \int_0^{\pi / 2} 1 d x = [x]_0^{\pi / 2} = \frac{\pi}{2}$. Thus, $I_1 = \frac{\pi}{4}$. Now, substituting back into the expression for $I$: $I = 100 	imes [x]_0^{\pi / 2} + 100 	imes I_1 = 100 	imes \frac{\pi}{2} + 100 	imes \frac{\pi}{4} = 50\pi + 25\pi = 75\pi$.

$\int_0^{\pi / 2} \frac{200 \sin x+100 \cos x}{\sin x+\cos x} d x$ is equal to (in $\pi$)

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