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amplitude of 2cos(3x)

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∫ cos(2x+5)dx

Solution

12 sin(2x+5)+C

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Solution steps

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∫ cos(2x+5)dx

Apply u-substitution: ∫ cos(u)12 du

=∫ cos(u)12 du

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

=12 · ∫ cos(u)du

Use the common integral: ∫ cos(u)du=sin(u)

=12 sin(u)

Substitute back u=2x+5

=12 sin(2x+5)

Add a constant to the solution

=12 sin(2x+5)+C

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

amplitude of 2cos(3x)

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