x^2+64

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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simplify solve for inverse tangent line

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∫ xcos(2x)dx

Solution

14 (2xsin(2x)+cos(2x))+C

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Solution steps

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∫ xcos(2x)dx

Apply u-substitution: ∫ ucos(u)4 du

=∫ ucos(u)4 du

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

=14 · ∫ ucos(u)du

Apply Integration By Parts: usin(u)−∫ sin(u)du

=14 (usin(u)−∫ sin(u)du)

∫ sin(u)du=−cos(u)

=14 (usin(u)−(−cos(u)))

Substitute back u=2x

=14 (2xsin(2x)−(−cos(2x)))

Simplify

=14 (2xsin(2x)+cos(2x))

Add a constant to the solution

=14 (2xsin(2x)+cos(2x))+C

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

x^2+64

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