What is the integral of 9/(θ^2)cos(9/θ) ?

The integral of 9/(θ^2)cos(9/θ) is -sin(9/θ)+C

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integral of 9/(θ^2)cos(9/θ)

\int \frac{9}{θ ^{2}} cos (\frac{9}{θ}) d θ

- sin (\frac{9}{θ}) + C

Solution steps

\int \frac{9}{θ ^{2}} cos (\frac{9}{θ}) d θ

Take the constant out:

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

= 9 \cdot \int \frac{1}{θ ^{2}} cos (\frac{9}{θ}) d θ

Apply u-substitution

: \int - \frac{cos ( u )}{9} d u

= 9 \cdot \int - \frac{cos ( u )}{9} d u

Take the constant out:

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

= 9 (- \frac{1}{9} \cdot \int cos (u) d u)

Use the common integral:

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

= 9 (- \frac{1}{9} sin (u))

Substitute back

u = \frac{9}{θ}

= 9 (- \frac{1}{9} sin (\frac{9}{θ}))

Simplify

9 (- \frac{1}{9} sin (\frac{9}{θ})) : - sin (\frac{9}{θ})

= - sin (\frac{9}{θ})

Add a constant to the solution

= - sin (\frac{9}{θ}) + C

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What is the integral of 9/(θ^2)cos(9/θ) ?
The integral of 9/(θ^2)cos(9/θ) is -sin(9/θ)+C