integral of (sin^3(x))/(cos^4(x))

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∫ sin³(x)cos⁴(x) dx

Solution

13 sec³(x)−sec(x)+C

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∫ sin³(x)cos⁴(x) dx

Rewrite using trig identities

=∫ (1−cos²(x))sin(x)cos⁴(x) dx

Apply u-substitution: ∫ −1−u²u⁴ du

=∫ −1−u²u⁴ du

Expand −1−u²u⁴ : −1u⁴ +1u²

=∫ −1u⁴ +1u² du

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

=−∫ 1u⁴ du+∫ 1u² du

∫ 1u⁴ du=−13u³

∫ 1u² du=−1u

=−(−13u³ )−1u

Substitute back u=cos(x)

=−(−13cos³(x) )−1cos(x)

Simplify −(−13cos³(x) )−1cos(x) : 13 sec³(x)−sec(x)

=13 sec³(x)−sec(x)

Add a constant to the solution

=13 sec³(x)−sec(x)+C

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

integral of (sin^3(x))/(cos^4(x))

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