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integral of sec^5(x)tan^5(x)

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

☐²

x^☐

√☐

^□√☐

□□

log_□

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∫

ddx

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

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développer (x−3)³

Solution

x³−9x²+27x−27

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(x−3)³

Appliquer la formule du cube parfait: (a−b)³=a³−3a²b+3ab²−b³

(x−3)³=x³−3x²· 3+3x· 3²−3³

=x³−3x²· 3+3x· 3²−3³

Simplifier x³−3x²· 3+3x· 3²−3³: x³−9x²+27x−27

=x³−9x²+27x−27

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

integral of sec^5(x)tan^5(x)

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