What is the integral of sin^2(8n-3) ?

The integral of sin^2(8n-3) is 1/16 (8n-3-1/2 sin(2(8n-3)))+C

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integral of sin^2(8n-3)

\int sin^{2} (8 n - 3) d n

\frac{1}{16} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3))) + C

Solution steps

\int sin^{2} (8 n - 3) d n

Apply u-substitution

: \int sin^{2} (u) \frac{1}{8} d u

= \int sin^{2} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{1}{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

= \frac{1}{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

= \frac{1}{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)

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

Substitute back

u = 8 n - 3

= \frac{1}{8} \cdot \frac{1}{2} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3)))

Simplify

\frac{1}{8} \cdot \frac{1}{2} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3))) : \frac{1}{16} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3)))

= \frac{1}{16} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3)))

Add a constant to the solution

= \frac{1}{16} (8 n - 3 - \frac{1}{2} sin (2 (8 n - 3))) + C

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What is the integral of sin^2(8n-3) ?
The integral of sin^2(8n-3) is 1/16 (8n-3-1/2 sin(2(8n-3)))+C