equation given points of (2,0), (3,-2), (1,-2)

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

☐²

x^☐

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log_□

θ

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∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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∫ e^xsin(3x)dx

Solution

−3e^xcos(3x)10 +e^xsin(3x)10 +C

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∫ e^xsin(3x)dx

Apply Integration By Parts: −13 e^xcos(3x)−∫ −13 e^xcos(3x)dx

=−13 e^xcos(3x)−∫ −13 e^xcos(3x)dx

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

=−13 e^xcos(3x)−(−13 · ∫ e^xcos(3x)dx)

Apply Integration By Parts: e^x13 sin(3x)−∫ e^x13 sin(3x)dx

=−13 e^xcos(3x)−(−13 (e^x13 sin(3x)−∫ e^x13 sin(3x)dx))

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

=−13 e^xcos(3x)−(−13 (e^x13 sin(3x)−13 · ∫ e^xsin(3x)dx))

Therefore

∫ e^xsin(3x)dx=−13 e^xcos(3x)−(−13 (e^x13 sin(3x)−13 · ∫ e^xsin(3x)dx))

Isolate ∫ e^xsin(3x)dx

=−3e^xcos(3x)10 +e^xsin(3x)10

Add a constant to the solution

=−3e^xcos(3x)10 +e^xsin(3x)10 +C

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

equation given points of (2,0), (3,-2), (1,-2)

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