I = ∫_0 ^(π/2) cos 2x(cos^4 x+sin^4 x) dx


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Question Number 87815 by jagoll last updated on 06/Apr/20

$I = \underset{0}{\int^{\frac{π}{2}}} \cos 2 x (\cos^{4} x + \sin^{4} x) dx$
Commented by john santu last updated on 06/Apr/20

$\cos^{4} x + \sin^{4} x = {(\cos^{2} x + \sin^{2} x)}^{2} - 2 \cos^{2} xsin^{2} x$ $= 1 - \frac{1}{2} \sin^{2} (2 x)$ $I = \int \frac{1}{2} \cos 2 x (2 - \sin^{2} (2 x)) dx$ $[let t = \sin 2 x \Rightarrow dt = 2 \cos 2 x dx]$ $I = \frac{1}{4} \int (2 - t^{2}) dt$ $I = \frac{1}{4} [2 t - \frac{1}{3} t^{3}]$ $now I = \frac{1}{4} {[2 \sin 2 x - \frac{1}{3} \sin^{3} 2 x]}_{0}^{\frac{π}{2}}$ $I = 0$
Commented by jagoll last updated on 06/Apr/20

$thank sir$
Commented by mathmax by abdo last updated on 06/Apr/20
$I =∫_0 ^(π/2) cos(2x)(cos^4 x +sin^4 x)dx changement x =(π/2)−t give I =−∫_0 ^(π/2) cos(π−2t){sin^4 t +cos^4 t)(−dt) =−∫_0 ^(π/2) cos(2t){cos^4 t +sin^4 t}dt =−I ⇒2I=0 ⇒I =0$
$I = \int_{0}^{\frac{π}{2}} c o s (2 x) ({c o s}^{4} x + {s i n}^{4} x) d x c h a n g e m e n t x = \frac{π}{2} - t g i v e$ $I = - \int_{0}^{\frac{π}{2}} c o s (π - 2 t) {{s i n}^{4} t + {c o s}^{4} t) (- d t)$ $= - \int_{0}^{\frac{π}{2}} c o s (2 t) {{c o s}^{4} t + {s i n}^{4} t} d t = - I \Rightarrow 2 I = 0 \Rightarrow I = 0$
	Answered by redmiiuser last updated on 06/Apr/20

	$\sin^{4} x + \cos^{4} x$ $= {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x . \cos^{2} x$ $= 1 - 2 . \frac{{(\sin 2 x)}^{2}}{4}$ $= 1 - \frac{1 - \cos 4 x}{4}$ $= \frac{3 + \cos 4 x}{4}$ $\int_{0}^{\frac{π}{2}} (\frac{3}{4} \cos 2 x . d x + \frac{1}{4} \cos 2 x . \cos 4 x . d x)$ $= \frac{3}{8} {[\sin 2 x]}_{0}^{\frac{π}{2}} + \frac{1}{8} {[\frac{s i n 6 x}{6}]}_{0}^{\frac{π}{2}} + \frac{1}{8} {[\frac{\sin 2 x}{2}]}_{0}^{\frac{π}{2}}$ $= 0$
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