int_{-frac{pi}{2}}^{frac{pi}{2}} sin^2 x cos^ | vedclass

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(C) माना $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx$ ...$(i)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ क

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$\frac{4}{15}$

Vedclass · Answer

(C) माना $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx$ ...$(i)$गुणधर्म $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ का उपयोग करने पर, जहाँ $a = -\frac{\pi}{2}$ और $b = \frac{\pi}{2}$, अतः $a+b = 0$.$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2(-x) \cos^2(-x)(\sin(-x) + \cos(-x)) dx$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(-\sin x + \cos x) dx$ ...(ii)$(i)$ और (ii) को जोड़ने पर:$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x - \sin x + \cos x) dx$$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(2 \cos x) dx$$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$चूंकि $\sin^2 x \cos^3 x$ एक सम फलन है, इसलिए $I = 2 \int_0^{\frac{\pi}{2}} \sin^2 x \cos^3 x dx$.माना $\sin x = t$, तब $\cos x dx = dt$. जब $x=0, t=0$; जब $x=\frac{\pi}{2}, t=1$.$I = 2 \int_0^1 t^2 (1-t^2) dt = 2 \int_0^1 (t^2 - t^4) dt$$I = 2 \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{5} \right) = 2 \left( \frac{2}{15} \right) = \frac{4}{15}$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x(\sin x + \cos x) dx =$

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