integration of sin cube 2x
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We know that,
sin 3θ =3 sin θ - 4 sin3θ
⇒4 sin3θ = 3 sin θ - sin 3θ
⇒sin3θ =( 3 sin θ - sin 3θ) /4
Let I = ∫sin3 2x dx
=1/4∫3 sin 2x - sin 6x=
1/4-3 cos 2x2 + cos 6x6 + C
sin 3θ =3 sin θ - 4 sin3θ
⇒4 sin3θ = 3 sin θ - sin 3θ
⇒sin3θ =( 3 sin θ - sin 3θ) /4
Let I = ∫sin3 2x dx
=1/4∫3 sin 2x - sin 6x=
1/4-3 cos 2x2 + cos 6x6 + C
cutiepie017:
fine
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