What is the integral of 8sin^2(t+pi/(12)) ?

The integral of 8sin^2(t+pi/(12)) is 4(t+pi/(12)-1/2 sin(2(t+pi/(12))))+C

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integral of 8sin^2(t+pi/(12))

\int 8 sin^{2} (t + \frac{π}{12}) d t

4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) + C

Solution steps

\int 8 sin^{2} (t + \frac{π}{12}) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int sin^{2} (t + \frac{π}{12}) d t

Apply u-substitution

: \int sin^{2} (u) d u

= 8 \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= 8 \cdot \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \frac{1}{2} \cdot \int 1 - cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 8 \cdot \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= 8 \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = t + \frac{π}{12}

= 8 \cdot \frac{1}{2} (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

Simplify

8 \cdot \frac{1}{2} (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) : 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

= 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12})))

Add a constant to the solution

= 4 (t + \frac{π}{12} - \frac{1}{2} sin (2 (t + \frac{π}{12}))) + C

x y^{^{'}} = y + 6 x^{2} sin (x), y (π) = 0

x \to - 1 lim (x^{2} - 3 x - 4)

P = x^{3} - 7 x P, (2, - 6)

x \to 4 lim (\frac{3}{( x - 4 ) ^{2}})

\int cos^{4} (\frac{1}{2} x) d x

What is the integral of 8sin^2(t+pi/(12)) ?
The integral of 8sin^2(t+pi/(12)) is 4(t+pi/(12)-1/2 sin(2(t+pi/(12))))+C