What is the integral of 4/(θ^2)cos(4/θ) ?

The integral of 4/(θ^2)cos(4/θ) is -sin(4/θ)+C

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integral of 4/(θ^2)cos(4/θ)

\int \frac{4}{θ ^{2}} cos (\frac{4}{θ}) d θ

- sin (\frac{4}{θ}) + C

Solution steps

\int \frac{4}{θ ^{2}} cos (\frac{4}{θ}) d θ

Take the constant out:

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

= 4 \cdot \int \frac{1}{θ ^{2}} cos (\frac{4}{θ}) d θ

Apply u-substitution

: \int - \frac{cos ( u )}{4} d u

= 4 \cdot \int - \frac{cos ( u )}{4} d u

Take the constant out:

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

= 4 (- \frac{1}{4} \cdot \int cos (u) d u)

Use the common integral:

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

= 4 (- \frac{1}{4} sin (u))

Substitute back

u = \frac{4}{θ}

= 4 (- \frac{1}{4} sin (\frac{4}{θ}))

Simplify

4 (- \frac{1}{4} sin (\frac{4}{θ})) : - sin (\frac{4}{θ})

= - sin (\frac{4}{θ})

Add a constant to the solution

= - sin (\frac{4}{θ}) + C

\int \frac{1}{x ( x - 4 )} d x

f (x) = \frac{2 t}{2 t}

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{x} )}{1 + x ^{2}}

\frac{d}{d t} (1 - t^{2})

\frac{d}{d x} (5 e^{x} - b)

What is the integral of 4/(θ^2)cos(4/θ) ?
The integral of 4/(θ^2)cos(4/θ) is -sin(4/θ)+C