integral of sin^2(t)

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∫ sin²(t)dt

Solution

12 (t−12 sin(2t))+C

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∫ sin²(t)dt

Rewrite using trig identities

=∫ 1−cos(2t)2 dt

Take the constant out: ∫a·f(x)dx=a·∫f(x)dx

=12 · ∫ 1−cos(2t)dt

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

=12 (∫ 1dt−∫ cos(2t)dt)

∫ 1dt=t

∫ cos(2t)dt=12 sin(2t)

=12 (t−12 sin(2t))

Add a constant to the solution

=12 (t−12 sin(2t))+C

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

integral of sin^2(t)

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