∫sen²(x) dx=
cos 2x=cos²x-sen²x
cos 2x=1-2sen²x
sen²x=(1-cos2x)/2
∫(1-cos2x)/2 dx=
1/2∫ dx -1/2∫cos 2x dx
x/2 -1/2∫ cos u dx
u=2x ==>du=2 dx
x/2 -1/2∫ cos u du/2
x/2 -1/4∫ cos u du
x/2 -1/4* sen u + c
x/2 - sen (2x) /4 +c
x/2 - sen x*cos x /2 +c
1/2(x-senx*cos x) +c
Segue resposta
1/2 (x - sin x * cos x) + cte
http://www.wolframalpha.com/input/?i=integral+sin%...
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Verified answer
∫sen²(x) dx=
cos 2x=cos²x-sen²x
cos 2x=1-2sen²x
sen²x=(1-cos2x)/2
∫(1-cos2x)/2 dx=
1/2∫ dx -1/2∫cos 2x dx
x/2 -1/2∫ cos u dx
u=2x ==>du=2 dx
x/2 -1/2∫ cos u du/2
x/2 -1/4∫ cos u du
x/2 -1/4* sen u + c
x/2 - sen (2x) /4 +c
x/2 - sen x*cos x /2 +c
1/2(x-senx*cos x) +c
Segue resposta
1/2 (x - sin x * cos x) + cte
http://www.wolframalpha.com/input/?i=integral+sin%...