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integral of tan^5(x)

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∫ tan⁵(x)dx

Solution

ln|sec(x)|−sec²(x)+sec⁴(x)4 +C

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∫ tan⁵(x)dx

Rewrite using trig identities

=∫ (−1+sec²(x))²tan(x)dx

Apply u-substitution: ∫ (−1+u²)²u du

=∫ (−1+u²)²u du

Expand (−1+u²)²u : 1u −2u+u³

=∫ 1u −2u+u³du

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

=∫ 1u du−∫ 2udu+∫ u³du

∫ 1u du=ln|u|

∫ 2udu=u²

∫ u³du=u⁴4

=ln|u|−u²+u⁴4

Substitute back u=sec(x)

=ln|sec(x)|−sec²(x)+sec⁴(x)4

Add a constant to the solution

=ln|sec(x)|−sec²(x)+sec⁴(x)4 +C

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

integral of tan^5(x)

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