If int_0^{2 pi}left(sin ^4 x+cos ^4 xright) d | vedclass

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(B) $L$.$H$.$S$. $= \int_0^{2 \pi} (\sin ^4 x + \cos ^4 x) dx = 2 \int_0^{\pi} (\sin ^4 x + \cos ^4 x) dx = 4 \int_0^{\pi/2} (\sin ^4 x + \cos ^4 x) d

Vedclass · Accepted Answer

$2$

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(B) $L$.$H$.$S$. $= \int_0^{2 \pi} (\sin ^4 x + \cos ^4 x) dx = 2 \int_0^{\pi} (\sin ^4 x + \cos ^4 x) dx = 4 \int_0^{\pi/2} (\sin ^4 x + \cos ^4 x) dx$. Using Wallis formula,$\int_0^{\pi/2} \sin^4 x dx = \int_0^{\pi/2} \cos^4 x dx = \frac{3 	imes 1}{4 	imes 2} 	imes \frac{\pi}{2} = \frac{3\pi}{16}$. So,$L$.$H$.$S$. $= 4 	imes (\frac{3\pi}{16} + \frac{3\pi}{16}) = 4 	imes \frac{6\pi}{16} = \frac{3\pi}{2}$. $R$.$H$.$S$. $= K \int_0^{\pi} \sin^2 x dx + L \int_0^{\pi/2} \cos^2 x dx = K(2 \int_0^{\pi/2} \sin^2 x dx) + L \int_0^{\pi/2} \cos^2 x dx$. Using $\int_0^{\pi/2} \sin^2 x dx = \int_0^{\pi/2} \cos^2 x dx = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4}$. $R$.$H$.$S$. $= K(2 	imes \frac{\pi}{4}) + L(\frac{\pi}{4}) = \frac{K\pi}{2} + \frac{L\pi}{4} = \frac{(2K + L)\pi}{4}$. Equating $L$.$H$.$S$. and $R$.$H$.$S$.: $\frac{3\pi}{2} = \frac{(2K + L)\pi}{4} \Rightarrow 6 = 2K + L$. Since $K, L \in N$ (Natural numbers),we test values for $K$: If $K=1$,$L = 6 - 2(1) = 4$. If $K=2$,$L = 6 - 2(2) = 2$. If $K=3$,$L = 6 - 2(3) = 0$ (Not in $N$). Thus,the possible ordered pairs $(K, L)$ are $(1, 4)$ and $(2, 2)$. There are $2$ such pairs.

If $\int_0^{2 \pi}\left(\sin ^4 x+\cos ^4 x\right) d x=K \int_0^\pi \sin ^2 x d x+L \int_0^{\frac{\pi}{2}} \cos ^2 x d x$ and $K, L \in N$,then the number of possible ordered pairs $(K, L)$ is

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