vertices x^2+1

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

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≥	≤	·	÷	`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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∫ cos³(x)sin⁸(x)dx

Solution

sin⁹(x)9 −sin¹¹(x)11 +C

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∫ cos³(x)sin⁸(x)dx

Rewrite using trig identities

=∫ (1−sin²(x))cos(x)sin⁸(x)dx

Apply u-substitution: ∫ u⁸(1−u²)du

=∫ u⁸(1−u²)du

Expand u⁸(1−u²): u⁸−u¹⁰

=∫ u⁸−u¹⁰du

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

=∫ u⁸du−∫ u¹⁰du

∫ u⁸du=u⁹9

∫ u¹⁰du=u¹¹11

=u⁹9 −u¹¹11

Substitute back u=sin(x)

=sin⁹(x)9 −sin¹¹(x)11

Add a constant to the solution

=sin⁹(x)9 −sin¹¹(x)11 +C

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

vertices x^2+1

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