find the values of I = ∫_0 ^π cos^4 dx and J = ∫


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Question Number 36406 by abdo mathsup 649 cc last updated on 01/Jun/18

$f i n d t h e v a l u e s o f I = \int_{0}^{π} {c o s}^{4} d x a n d$ $J = \int_{0}^{π} {s i n}^{4} d x .$
Commented by abdo.msup.com last updated on 05/Jun/18
$we have I +J =∫_0 ^π (cos^4 x +sin^4 x)dx =∫_0 ^π {(cos^2 x +sin^2 x)^2 −2cos^2 xsin^2 x}dx =π −2 ∫_0 ^π (1/4)(sin(2x))^2 dx =π −(1/2) ∫_0 ^π ((1−cos(4x))/2)dx =π −(π/4) + (1/(16))[sin(4x)]_0 ^π =((3π)/4) also I −J = ∫_0 ^π (cos^4 x −sin^4 x)dx = ∫_0 ^π cos^2 x −sin^2 x dx = ∫_0 ^π cos(2x)dx=(1/2)[ sin(2x)]_0 ^π =0 so I =J ⇒ 2I = ((3π)/4) ⇒ I =((3π)/8) and J =((3π)/8)$
$w e h a v e I + J = \int_{0}^{π} ({c o s}^{4} x + {s i n}^{4} x) d x$ $= \int_{0}^{π} {{({c o s}^{2} x + {s i n}^{2} x)}^{2} - 2 {c o s}^{2} {x s i n}^{2} x} d x$ $= π - 2 \int_{0}^{π} \frac{1}{4} {(s i n (2 x))}^{2} d x$ $= π - \frac{1}{2} \int_{0}^{π} \frac{1 - c o s (4 x)}{2} d x$ $= π - \frac{π}{4} + \frac{1}{16} {[s i n (4 x)]}_{0}^{π} = \frac{3 π}{4} a l s o$ $I - J = \int_{0}^{π} ({c o s}^{4} x - {s i n}^{4} x) d x$ $= \int_{0}^{π} {c o s}^{2} x - {s i n}^{2} x d x$ $= \int_{0}^{π} c o s (2 x) d x = \frac{1}{2} {[s i n (2 x)]}_{0}^{π} = 0 s o$ $I = J \Rightarrow 2 I = \frac{3 π}{4} \Rightarrow I = \frac{3 π}{8} a n d J = \frac{3 π}{8}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

	$= \frac{1}{4} \int_{0}^{Π} {(1 + c o s 2 x)}^{2} d x$ $= \frac{1}{4} \int_{0}^{Π} (1 + 2 c o s 2 x + {c o s}^{2} 2 x) d x$ $= \frac{1}{4} \int_{0}^{Π} d x + \frac{1}{2} \int_{0}^{Π} c o s 2 x d x + \frac{1}{8} \int_{0}^{Π} (1 + c o s 4 x) d x$ $= \frac{1}{4} \int_{0}^{Π} d x + \frac{1}{2} \int_{0}^{Π} c o s 2 x d x + \frac{1}{8} \int_{0}^{Π} d x + \frac{1}{8} \int_{0}^{Π} c o s 4 x$ $= \frac{Π}{4} + \frac{Π}{8}$ $= \frac{3 Π}{8}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

	$j = \int_{0}^{Π (} {(\frac{1 - c o s 2 x}{2})}^{2}$ $= \frac{1}{4} \int_{0}^{Π} 1 - 2 c o s 2 x + \frac{1 + c o s 4 x}{2} d x$ $= \frac{1}{4} \int_{0}^{Π} d x - \frac{1}{2} \int_{0}^{Π} c o s 2 x d x + \frac{1}{8} \int_{0}^{Π} d x + \frac{1}{8} \int_{0}^{Π} c o s 4 x$ $= \frac{Π}{4} + \frac{Π}{8} = \frac{3 Π}{8}$
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