int_0^pi (sin^3 x + cos^2 x)^2 dx = | vedclass

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(D) Let $I = \int_0^\pi (\sin^3 x + \cos^2 x)^2 dx$. Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$, we note that $

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$\frac{11\pi}{16} + \frac{4}{15}$

Vedclass · Answer

(D) Let $I = \int_0^\pi (\sin^3 x + \cos^2 x)^2 dx$. Using the property $\int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx$ if $f(2a-x) = f(x)$, we note that $(\sin^3(\pi-x) + \cos^2(\pi-x))^2 = (\sin^3 x + \cos^2 x)^2$. Thus, $I = 2 \int_0^{\pi/2} (\sin^6 x + \cos^4 x + 2 \sin^3 x \cos^2 x) dx$. Using Wallis' formula $\int_0^{\pi/2} \sin^n x dx = \int_0^{\pi/2} \cos^n x dx = \frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ (for even $n$) or $\frac{(n-1)!!}{n!!}$ (for odd $n$): $I = 2 [ \int_0^{\pi/2} \sin^6 x dx + \int_0^{\pi/2} \cos^4 x dx + 2 \int_0^{\pi/2} \sin^3 x \cos^2 x dx ]$. $I = 2 [ (\frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}) + (\frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2}) + 2 \int_0^{\pi/2} (1-\cos^2 x) \cos^2 x \sin x dx ]$. Let $u = \cos x$, then $du = -\sin x dx$. $2 \int_0^{\pi/2} (1-\cos^2 x) \cos^2 x \sin x dx = 2 \int_0^1 (u^2 - u^4) du = 2 [\frac{u^3}{3} - \frac{u^5}{5}]_0^1 = 2(\frac{1}{3} - \frac{1}{5}) = 2(\frac{2}{15}) = \frac{4}{15}$. $I = 2 [ \frac{5\pi}{32} + \frac{3\pi}{16} ] + \frac{4}{15} = 2 [ \frac{5\pi + 6\pi}{32} ] + \frac{4}{15} = \frac{11\pi}{16} + \frac{4}{15}$.

$\int_0^\pi (\sin^3 x + \cos^2 x)^2 dx = $

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