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

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(D) ધારો કે $I = \int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx$. આપણે ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉ

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) ધારો કે $I = \int_0^\pi (\sin^5 x \cos^3 x + \sin^4 x \cos^4 x + \sin^3 x \cos^4 x) dx$. આપણે ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરીએ છીએ. $f_1(x) = \sin^5 x \cos^3 x$ માટે,$f_1(\pi-x) = \sin^5 x (-\cos x)^3 = -\sin^5 x \cos^3 x$. તેથી,$\int_0^\pi \sin^5 x \cos^3 x dx = 0$. $f_3(x) = \sin^3 x \cos^4 x$ માટે,$f_3(\pi-x) = \sin^3 x (-\cos x)^4 = \sin^3 x \cos^4 x$. $\int_0^\pi f(x) dx = 2 \int_0^{\pi/2} f(x) dx$ નો ઉપયોગ કરતા,$\int_0^\pi \sin^3 x \cos^4 x dx = 2 \int_0^{\pi/2} \sin^3 x \cos^4 x dx = 2 \cdot \frac{\Gamma(2) \Gamma(5/2)}{2 \Gamma(9/2)} = \frac{4}{35}$. $f_2(x) = \sin^4 x \cos^4 x = (\frac{1}{2} \sin 2x)^4 = \frac{1}{16} \sin^4 2x$ માટે. $\int_0^\pi \frac{1}{16} \sin^4 2x dx = \frac{3\pi}{128}$. સરવાળો કરતા,$I = 0 + \frac{3\pi}{128} + \frac{4}{35} = \frac{3\pi}{128} + \frac{4}{35}$.

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

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