integral of sec^3(theta)tan^3(theta)heta

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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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∫ sec³(θ)tan³(θ)dθ

Solution

sec⁵(θ)5 −sec³(θ)3 +C

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∫ sec³(θ)tan³(θ)dθ

Rewrite using trig identities

=∫ (−1+sec²(θ))tan(θ)sec³(θ)dθ

Apply u-substitution: ∫ u²(u²−1)du

=∫ u²(u²−1)du

Expand u²(u²−1): u⁴−u²

=∫ u⁴−u²du

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

=∫ u⁴du−∫ u²du

∫ u⁴du=u⁵5

∫ u²du=u³3

=u⁵5 −u³3

Substitute back u=sec(θ)

=sec⁵(θ)5 −sec³(θ)3

Add a constant to the solution

=sec⁵(θ)5 −sec³(θ)3 +C

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

integral of sec^3(theta)tan^3(theta)heta

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