What is the integral of 1/(t^2)sin(3/t+6) ?

The integral of 1/(t^2)sin(3/t+6) is 1/3 cos(3/t+6)+C

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integral of 1/(t^2)sin(3/t+6)

\int \frac{1}{t ^{2}} sin (\frac{3}{t} + 6) d t

\frac{1}{3} cos (\frac{3}{t} + 6) + C

Solution steps

\int \frac{1}{t ^{2}} sin (\frac{3}{t} + 6) d t

Simplify

\frac{1}{t ^{2}} sin (\frac{3}{t} + 6) : \frac{sin ( \frac{3}{t} + 6 )}{t ^{2}}

= \int \frac{sin ( \frac{3}{t} + 6 )}{t ^{2}} d t

Apply u-substitution

: \int - \frac{sin ( u )}{3} d u

= \int - \frac{sin ( u )}{3} d u

Take the constant out:

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

= - \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= - \frac{1}{3} (- cos (u))

Substitute back

u = \frac{3}{t} + 6

= - \frac{1}{3} (- cos (\frac{3}{t} + 6))

Simplify

= \frac{1}{3} cos (\frac{3}{t} + 6)

Add a constant to the solution

= \frac{1}{3} cos (\frac{3}{t} + 6) + C

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What is the integral of 1/(t^2)sin(3/t+6) ?
The integral of 1/(t^2)sin(3/t+6) is 1/3 cos(3/t+6)+C