What is the integral of 1/(t^2)sin(8/t+3) ?

The integral of 1/(t^2)sin(8/t+3) is 1/8 cos(8/t+3)+C

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integral of 1/(t^2)sin(8/t+3)

\int \frac{1}{t ^{2}} sin (\frac{8}{t} + 3) d t

\frac{1}{8} cos (\frac{8}{t} + 3) + C

Solution steps

\int \frac{1}{t ^{2}} sin (\frac{8}{t} + 3) d t

Simplify

\frac{1}{t ^{2}} sin (\frac{8}{t} + 3) : \frac{sin ( \frac{8}{t} + 3 )}{t ^{2}}

= \int \frac{sin ( \frac{8}{t} + 3 )}{t ^{2}} d t

Apply u-substitution

: \int - \frac{sin ( u )}{8} d u

= \int - \frac{sin ( u )}{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 (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= - \frac{1}{8} (- cos (u))

Substitute back

u = \frac{8}{t} + 3

= - \frac{1}{8} (- cos (\frac{8}{t} + 3))

Simplify

= \frac{1}{8} cos (\frac{8}{t} + 3)

Add a constant to the solution

= \frac{1}{8} cos (\frac{8}{t} + 3) + C

3 + \frac{1}{s}

\int_{o}^{1} 42 x (1 - x)^{5} d x

\frac{d}{d x} (a^{2} + x^{2})

x \to 1 lim (\frac{1 + x}{1 - x ^{2}})

\int_{0}^{- \frac{π}{6}} x cos (3 x) d x

What is the integral of 1/(t^2)sin(8/t+3) ?
The integral of 1/(t^2)sin(8/t+3) is 1/8 cos(8/t+3)+C