calculate ∫_0 ^(π/4) cos^4 x sin^2 xdx


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Question Number 40130 by maxmathsup by imad last updated on 16/Jul/18

$c a l c u l a t e \int_{0}^{\frac{π}{4}} {c o s}^{4} x {s i n}^{2} x d x$
Commented by math khazana by abdo last updated on 21/Jul/18
$let I = ∫_0 ^(π/4) cos^4 x sin^2 xdx by parts I = ∫_0 ^(π/4) sinx cos^4 x sinx dx =[−(1/5) cos^5 x sinx]_0 ^(π/4) −∫_0 ^(π/4) (−(1/5))cos^5 x cosxdx =−(1/5)(((√2)/2))^6 +(1/5) ∫_0 ^(π/4) cos^6 x dx but ∫_0 ^(π/4) cos^6 xdx = ∫_0 ^(π/4) (((1+cos(2x))/2))^3 dx =(1/8) ∫_0 ^(π/4) { cos^3 (2x) +3cos^2 (2x) +3cos(2x)+1)}dx =(1/8) ∫_0 ^(π/4) cos^3 (2x)dx +(3/8) ∫_0 ^(π/4) ((1+cos(4x))/2)dx +(3/8) ∫_0 ^(π/4) cos(2x)dx +(π/(32)) but ∫_0 ^(π/4) cos(2x)dx=[(1/2)sin(2x)]_0 ^(π/4) =(1/2) ∫_0 ^(π/4) ((1+cos(4x))/2)dx =(π/8) +(1/8)[sin(4x)]_0 ^(π/4) =(π/8) ∫_0 ^(π/4) cos^3 (2x)dx = ∫_0 ^(π/4) cos(2x)((1+cos(4x))/2)dx =(1/2) ∫_0 ^(π/4) cos(2x) +(1/2) ∫_0 ^(π/4) cos(2x)cos(4x)dx =(1/4)[sin(2x)]_0 ^(π/4) +(1/4) ∫_0 ^(π/4) {cos(6x)+cos(2x))dx =(1/4) +(1/(24))[sin(6x)]_0 ^(π/4) +(1/8)[sin(2x)]_0 ^(π/4) =(1/4) −(1/(24)) +(1/8) =((6−1+3)/(24)) =(8/(24)) =(1/3) so the value of I is determined.$
$l e t I = \int_{0}^{\frac{π}{4}} {c o s}^{4} x {s i n}^{2} x d x b y p a r t s$ $I = \int_{0}^{\frac{π}{4}} s i n x {c o s}^{4} x s i n x d x$ $= {[- \frac{1}{5} {c o s}^{5} x s i n x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} (- \frac{1}{5}) {c o s}^{5} x c o s x d x$ $= - \frac{1}{5} {(\frac{\sqrt{2}}{2})}^{6} + \frac{1}{5} \int_{0}^{\frac{π}{4}} {c o s}^{6} x d x b u t$ $\int_{0}^{\frac{π}{4}} {c o s}^{6} x d x = \int_{0}^{\frac{π}{4}} {(\frac{1 + c o s (2 x)}{2})}^{3} d x$ $= \frac{1}{8} \int_{0}^{\frac{π}{4}} {{c o s}^{3} (2 x) + 3 {c o s}^{2} (2 x) + 3 c o s (2 x) + 1)} d x$ $= \frac{1}{8} \int_{0}^{\frac{π}{4}} {c o s}^{3} (2 x) d x + \frac{3}{8} \int_{0}^{\frac{π}{4}} \frac{1 + c o s (4 x)}{2} d x$ $+ \frac{3}{8} \int_{0}^{\frac{π}{4}} c o s (2 x) d x + \frac{π}{32} b u t$ $\int_{0}^{\frac{π}{4}} c o s (2 x) d x = {[\frac{1}{2} s i n (2 x)]}_{0}^{\frac{π}{4}} = \frac{1}{2}$ $\int_{0}^{\frac{π}{4}} \frac{1 + c o s (4 x)}{2} d x = \frac{π}{8} + \frac{1}{8} {[s i n (4 x)]}_{0}^{\frac{π}{4}}$ $= \frac{π}{8}$ $\int_{0}^{\frac{π}{4}} {c o s}^{3} (2 x) d x = \int_{0}^{\frac{π}{4}} c o s (2 x) \frac{1 + c o s (4 x)}{2} d x$ $= \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 x) + \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 x) c o s (4 x) d x$ $= \frac{1}{4} {[s i n (2 x)]}_{0}^{\frac{π}{4}} + \frac{1}{4} \int_{0}^{\frac{π}{4}} {c o s (6 x) + c o s (2 x)) d x$ $= \frac{1}{4} + \frac{1}{24} {[s i n (6 x)]}_{0}^{\frac{π}{4}} + \frac{1}{8} {[s i n (2 x)]}_{0}^{\frac{π}{4}}$ $= \frac{1}{4} - \frac{1}{24} + \frac{1}{8} = \frac{6 - 1 + 3}{24} = \frac{8}{24} = \frac{1}{3}$ $s o t h e v a l u e o f I i s d e t e r m i n e d .$
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