∫_(π/2) ^(π/4) sin^3 (x)cos^2 (x)dx


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Question Number 154877 by SANOGO last updated on 22/Sep/21

$\int_{\frac{π}{2}}^{\frac{π}{4}} {s i n}^{3} (x) {c o s}^{2} (x) d x$
Commented by saly last updated on 24/Sep/21

$t h a n k y o u$
	Answered by Mr.D.N. last updated on 22/Sep/21
	$∫_(π/2) ^( (π/4)) sin^3 (x) cos^2 (x) dx ∫_(π/2) ^( (π/4)) sin^2 x.cos^2 x. sin x dx ∫_(π/2) ^( (π/4)) (1−cos^2 x)cos^2 x sinx dx put cos x= t −sin x dx =dt sin x dx= −dt change limit, if x=(π/4) then t= (1/( (√2))) if x= (π/2) then t= 0 −∫_0 ^( (1/( (√2)))) (1−t^2 )t^2 dt ∫_0 ^(1/( (√2))) ( t^4 −t^2 )dt = [(t^5 /5)−(t^3 /3)]_0 ^(1/( (√2))) = [t^3 ((1/5)t^2 −(1/3))]_0 ^(1/(√2)) = [ ((1/( (√2))))^3 {(1/5)((√2) )^2 −(1/3)}]−0 = (1/( (√2)))×(1/2)((2/5)−(1/3)) = (1/( 2(√2)))(((6−5)/(15)))=(1/(30(√2))) //.$
	$\int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{3} (x) \cos^{2} (x) dx$ $\int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{2} x . \cos^{2} x . \sin x dx$ $\int_{\frac{π}{2}}^{\frac{π}{4}} (1 - \cos^{2} x) \cos^{2} x sinx dx$ $put \cos x = t$ $- \sin x dx = dt$ $\sin x dx = - dt$ $change limit,$ $if x = \frac{π}{4} then t = \frac{1}{\sqrt{2}}$ $if x = \frac{π}{2} then t = 0$ $- \int_{0}^{\frac{1}{\sqrt{2}}} (1 - t^{2}) t^{2} dt$ $\int_{0}^{\frac{1}{\sqrt{2}}} (t^{4} - t^{2}) dt$ $= {[\frac{t^{5}}{5} - \frac{t^{3}}{3}]}_{0}^{\frac{1}{\sqrt{2}}}$ $= {[t^{3} (\frac{1}{5} t^{2} - \frac{1}{3})]}_{0}^{1 / \sqrt{2}}$ $= [{(\frac{1}{\sqrt{2}})}^{3} {\frac{1}{5} {(\sqrt{2})}^{2} - \frac{1}{3}}] - 0$ $= \frac{1}{\sqrt{2}} \times \frac{1}{2} (\frac{2}{5} - \frac{1}{3})$ $= \frac{1}{2 \sqrt{2}} (\frac{6 - 5}{15}) = \frac{1}{30 \sqrt{2}} / / .$
	Commented by SANOGO last updated on 23/Sep/21

	$m e r c i b i e n$
	Answered by Ar Brandon last updated on 23/Sep/21

	$I = \int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{3} x \cos^{2} x d x = \int_{\frac{π}{2}}^{\frac{π}{4}} \sin x (\cos^{2} x - \cos^{4} x) d x$ $= \int_{\frac{π}{2}}^{\frac{π}{4}} (\cos^{4} x - \cos^{2} x) d (\cos x) = {[\frac{\cos^{5} x}{5} - \frac{\cos^{3} x}{3}]}_{\frac{π}{2}}^{\frac{π}{4}}$ $= \frac{1}{5 \sqrt{2^{5}}} - \frac{1}{3 \sqrt{2^{3}}} = \frac{1}{20 \sqrt{2}} - \frac{1}{6 \sqrt{2}} = - \frac{7 \sqrt{2}}{120}$
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