What is the integral of 8/(θ^2)cos(8/θ) ?

The integral of 8/(θ^2)cos(8/θ) is -sin(8/θ)+C

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integral of 8/(θ^2)cos(8/θ)

\int \frac{8}{θ ^{2}} cos (\frac{8}{θ}) d θ

- sin (\frac{8}{θ}) + C

Solution steps

\int \frac{8}{θ ^{2}} cos (\frac{8}{θ}) d θ

Take the constant out:

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

= 8 \cdot \int \frac{1}{θ ^{2}} cos (\frac{8}{θ}) d θ

Apply u-substitution

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

= 8 \cdot \int - \frac{cos ( u )}{8} d u

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

u = \frac{8}{θ}

= 8 (- \frac{1}{8} sin (\frac{8}{θ}))

Simplify

8 (- \frac{1}{8} sin (\frac{8}{θ})) : - sin (\frac{8}{θ})

= - sin (\frac{8}{θ})

Add a constant to the solution

= - sin (\frac{8}{θ}) + C

y = ln (8 - 4 x)

f (x) = 21 x

y = 8 x sin (x), (\frac{π}{2}, 4 π)

y = \frac{1}{x ^{5}}, (1, 1)

\frac{\partial}{\partial x} ((x_{1} + y_{2})^{2})

What is the integral of 8/(θ^2)cos(8/θ) ?
The integral of 8/(θ^2)cos(8/θ) is -sin(8/θ)+C