What is the integral of sin^2(x+4) ?

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

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integral of sin^2(x+4)

\int sin^{2} (x + 4) d x

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

Solution steps

\int sin^{2} (x + 4) d x

Apply u-substitution

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

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

Rewrite using trig identities

= \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

= \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

= \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)

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

Substitute back

u = x + 4

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

Add a constant to the solution

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

\int_{1}^{\infty} x e^{- 4 x} d x

\int sin^{4} (4 θ) d θ

\int (\frac{16}{1 + x ^{2}}) d x

sin (x^{2})

\frac{d}{d x} (x \cdot e^{4 x})

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