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integral of 2cos^3(x)

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∫ 2cos³(x)dx

Solution

2(sin(x)−sin³(x)3 )+C

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∫ 2cos³(x)dx

Take the constant out: ∫a·f(x)dx=a·∫f(x)dx

=2· ∫ cos³(x)dx

Rewrite using trig identities

=2· ∫ (1−sin²(x))cos(x)dx

Apply u-substitution: ∫ 1−u²du

=2· ∫ 1−u²du

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=2(∫ 1du−∫ u²du)

∫ 1du=u

∫ u²du=u³3

=2(u−u³3 )

Substitute back u=sin(x)

=2(sin(x)−sin³(x)3 )

Add a constant to the solution

=2(sin(x)−sin³(x)3 )+C

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

integral of 2cos^3(x)

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