(2^3)^3

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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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∫ sec⁵(x)tan³(x)dx

Solution

sec⁷(x)7 −sec⁵(x)5 +C

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Solution steps

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∫ sec⁵(x)tan³(x)dx

Rewrite using trig identities

=∫ (−1+sec²(x))tan(x)sec⁵(x)dx

Apply u-substitution: ∫ u⁴(u²−1)du

=∫ u⁴(u²−1)du

Expand u⁴(u²−1): u⁶−u⁴

=∫ u⁶−u⁴du

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

=∫ u⁶du−∫ u⁴du

∫ u⁶du=u⁷7

∫ u⁴du=u⁵5

=u⁷7 −u⁵5

Substitute back u=sec(x)

=sec⁷(x)7 −sec⁵(x)5

Add a constant to the solution

=sec⁷(x)7 −sec⁵(x)5 +C

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

(2^3)^3

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