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

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

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

\int 8 sin^{2} (x + \frac{π}{12}) d x

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

Solution steps

\int 8 sin^{2} (x + \frac{π}{12}) d x

Take the constant out:

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

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

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 = x + \frac{π}{12}

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

Simplify

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

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

Add a constant to the solution

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

n = 1 \sum \infty \infty

\int \frac{1}{x ^{\frac{3}{2}} + x ^{\frac{1}{2}}} d x

f (x) = \frac{x - 1}{x ^{2}}

\frac{\partial}{\partial x} (3 x^{2} y + 2 x y^{3})

n = 0 \sum \infty e^{- 1}

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