int_0^pi (sin^5 x cos^3 x + sin^4 x cos^4 x + | vedclass

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(D) Let $I = \int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx$. We use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$.

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$\frac{3\pi}{128} + \frac{4}{35}$

Vedclass · Answer

(D) Let $I = \int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx$. We use the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. For $f_1(x) = \sin^5 x \cos^3 x$,$f_1(\pi-x) = \sin^5 x (-\cos x)^3 = -\sin^5 x \cos^3 x$. Thus,$\int_0^\pi \sin^5 x \cos^3 x dx = 0$. For $f_3(x) = \sin^3 x \cos^4 x$,$f_3(\pi-x) = \sin^3 x (-\cos x)^4 = \sin^3 x \cos^4 x$. Using $\int_0^\pi f(x) dx = 2 \int_0^{\pi/2} f(x) dx$ for symmetric functions,$\int_0^\pi \sin^3 x \cos^4 x dx = 2 \int_0^{\pi/2} \sin^3 x \cos^4 x dx = 2 \cdot \frac{\Gamma(2) \Gamma(5/2)}{2 \Gamma(9/2)} = 2 \cdot \frac{1 \cdot (3/4 \cdot 1/2 \cdot \sqrt{\pi})}{7/2 \cdot 5/2 \cdot 3/2 \cdot 1/2 \cdot \sqrt{\pi}} = 2 \cdot \frac{3/8}{105/16} = 2 \cdot \frac{3}{8} \cdot \frac{16}{105} = \frac{4}{35}$. For $f_2(x) = \sin^4 x \cos^4 x = (\frac{1}{2} \sin 2x)^4 = \frac{1}{16} \sin^4 2x$. $\int_0^\pi \frac{1}{16} \sin^4 2x dx = \frac{1}{16} \cdot 2 \int_0^{\pi/2} \sin^4 2x dx$. Let $2x = t$,$dx = dt/2$. $= \frac{1}{8} \cdot \frac{1}{2} \int_0^{\pi} \sin^4 t dt = \frac{1}{16} \cdot 2 \int_0^{\pi/2} \sin^4 t dt = \frac{1}{8} \cdot \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{128}$. Summing these,$I = 0 + \frac{3\pi}{128} + \frac{4}{35} = \frac{3\pi}{128} + \frac{4}{35}$.

$\int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx =$

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