intigration of cos^3(3x+5)dx
Answers
Answer:
=sin(9x+15)+9sin(3x+5)36+C
Step-by-step explanation;
Problem:
∫cos3(3x+5)dx
Substitute u=3x+5 ⟶ dx=13du (steps):
=13∫cos3(u)du
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Now solving:
∫cos3(u)du
Prepare for substitution:
=∫cos(u)(1−sin2(u))du
Substitute v=sin(u) ⟶ du=1cos(u)dv (steps):
=∫(1−v2)dv
Apply linearity:
=∫1dv−∫v2dv
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Now solving:
∫1dv
Apply constant rule:
=v
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Now solving:
∫v2dv
Apply power rule:
∫vndv=vn+1n+1 with n=2:
=v33
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Plug in solved integrals:
∫1dv−∫v2dv
=v−v33
Undo substitution v=sin(u):
=sin(u)−sin3(u)3
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Plug in solved integrals:
13∫cos3(u)du
=sin(u)3−sin3(u)9
Undo substitution u=3x+5:
=sin(3x+5)3−sin3(3x+5)9
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The problem is solved:
∫cos3(3x+5)dx
=sin(3x+5)3−sin3(3x+5)9+C
Rewrite/simplify:
=sin(9x+15)+9sin(3x+5)36+C