Integrate sin^43x dx

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asked Dec 7, 2017 163k views

1 Answer

Ethan Waldie · Answer 1 · 2017-12-11T14:49:35+0000

sstarica the question is∫sin4 3x dx
substitute (cos x)^2 = 1 - (sin x)^2

∫(1−cos^2(3x))^2 dx = ∫(1−2cos^2u + cos^2u) du

1/15 ∫ 5(3)sin^4 (3x) dx

= -cos^5 (3x)/15 + C

Integrate sin^43x dx

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