periodizität sin^2(x)

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

☐²

x^☐

√☐

^□√☐

□□

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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domain sin(2π)dx

Solution

−∞ <x<∞

Interval Notation

(−∞ , ∞ )

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Solution steps

One step at a time

∫ 9−√81−5xx dx

Expand 9−√81−5xx : 9x −√81−5xx

=∫ 9x −√81−5xx dx

Apply the Sum Rule: ∫f(x)±g(x)dx=∫f(x)dx±∫g(x)dx

=∫ 9x dx−∫ √81−5xx dx

∫ 9x dx=9ln|x|

∫ √81−5xx dx=2√81−5x−9ln|19 √−5x+81+1|+9ln|19 √−5x+81−1|

=9ln|x|−(2√81−5x−9ln|19 √−5x+81+1|+9ln|19 √−5x+81−1|)

Simplify

=9ln|x|−2√81−5x+9ln|19 √−5x+81+1|−9ln|19 √−5x+81−1|

Add a constant to the solution

=9ln|x|−2√81−5x+9ln|19 √−5x+81+1|−9ln|19 √−5x+81−1|+C

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range sin(2π)dx

periodicity sin(2π)dx

inverse sin(2π)dx

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

periodizität sin^2(x)

Solution

Solution steps

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