vertices x^2+4

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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_□

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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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∫ sin⁷(x)cos⁴(x)dx

Solution

−cos⁵(x)5 +3cos⁷(x)7 −cos⁹(x)3 +cos¹¹(x)11 +C

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Solution steps

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∫ sin⁷(x)cos⁴(x)dx

Rewrite using trig identities

=∫ (1−cos²(x))³sin(x)cos⁴(x)dx

Apply u-substitution: ∫ −u⁴(1−u²)³du

=∫ −u⁴(1−u²)³du

Expand −u⁴(1−u²)³: −u⁴+3u⁶−3u⁸+u¹⁰

=∫ −u⁴+3u⁶−3u⁸+u¹⁰du

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

=−∫ u⁴du+∫ 3u⁶du−∫ 3u⁸du+∫ u¹⁰du

∫ u⁴du=u⁵5

∫ 3u⁶du=3u⁷7

∫ 3u⁸du=u⁹3

∫ u¹⁰du=u¹¹11

=−u⁵5 +3u⁷7 −u⁹3 +u¹¹11

Substitute back u=cos(x)

=−cos⁵(x)5 +3cos⁷(x)7 −cos⁹(x)3 +cos¹¹(x)11

Add a constant to the solution

=−cos⁵(x)5 +3cos⁷(x)7 −cos⁹(x)3 +cos¹¹(x)11 +C

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

vertices x^2+4

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