taylor 2cos^3(x)

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

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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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∫ √4x−3dx

Solution

16 (4x−3)³²+C

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∫ √4x−3dx

Apply u-substitution: ∫ √u4 du

=∫ √u4 du

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

=14 · ∫ √udu

Apply the Power Rule: 23 u³²

=14 · 23 u³²

Substitute back u=4x−3

=14 · 23 (4x−3)³²

Simplify 14 · 23 (4x−3)³²: 16 (4x−3)³²

=16 (4x−3)³²

Add a constant to the solution

=16 (4x−3)³²+C

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

taylor 2cos^3(x)

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