integral of tan^3(x)sec^4(x)

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∫ tan³(x)sec⁴(x)dx

Solution

tan⁴(x)4 +tan⁶(x)6 +C

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∫ tan³(x)sec⁴(x)dx

Rewrite using trig identities

=∫ (1+tan²(x))sec²(x)tan³(x)dx

Apply u-substitution: ∫ u³(1+u²)du

=∫ u³(1+u²)du

Expand u³(1+u²): u³+u⁵

=∫ u³+u⁵du

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

=∫ u³du+∫ u⁵du

∫ u³du=u⁴4

∫ u⁵du=u⁶6

=u⁴4 +u⁶6

Substitute back u=tan(x)

=tan⁴(x)4 +tan⁶(x)6

Add a constant to the solution

=tan⁴(x)4 +tan⁶(x)6 +C

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

integral of tan^3(x)sec^4(x)

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