What is the integral of 1/(t^3)sin(t^{-2}-9) ?

The integral of 1/(t^3)sin(t^{-2}-9) is 1/2 cos(1/(t^2)-9)+C

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integral of 1/(t^3)sin(t^{-2}-9)

\int \frac{1}{t ^{3}} sin (t^{- 2} - 9) d t

\frac{1}{2} cos (\frac{1}{t ^{2}} - 9) + C

Solution steps

\int \frac{1}{t ^{3}} sin (t^{- 2} - 9) d t

Apply u-substitution

: \int - \frac{sin ( u )}{2} d u

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

Use the common integral:

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

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

Substitute back

u = t^{- 2} - 9

= - \frac{1}{2} (- cos (t^{- 2} - 9))

Simplify

- \frac{1}{2} (- cos (t^{- 2} - 9)) : \frac{1}{2} cos (\frac{1}{t ^{2}} - 9)

= \frac{1}{2} cos (\frac{1}{t ^{2}} - 9)

Add a constant to the solution

= \frac{1}{2} cos (\frac{1}{t ^{2}} - 9) + C

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What is the integral of 1/(t^3)sin(t^{-2}-9) ?
The integral of 1/(t^3)sin(t^{-2}-9) is 1/2 cos(1/(t^2)-9)+C