Calculus Examples

$\frac{d y}{d θ} = \frac{1}{θ^{2}} - \csc (θ) \cot (θ)$

Step 1

Rewrite the equation.

$d y = (\frac{1}{θ^{2}} - \csc (θ) \cot (θ)) d θ$

Step 2

Integrate both sides.

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Step 2.1

Set up an integral on each side.

$\int d y = \int \frac{1}{θ^{2}} - \csc (θ) \cot (θ) d θ$

Step 2.2

Apply the constant rule.

$y + C_{1} = \int \frac{1}{θ^{2}} - \csc (θ) \cot (θ) d θ$

Step 2.3

Integrate the right side.

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Step 2.3.1

Split the single integral into multiple integrals.

$y + C_{1} = \int \frac{1}{θ^{2}} d θ + \int - \csc (θ) \cot (θ) d θ$

Step 2.3.2

Apply basic rules of exponents.

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Step 2.3.2.1

Move $θ^{2}$ out of the denominator by raising it to the $- 1$ power.

$y + C_{1} = \int {(θ^{2})}^{- 1} d θ + \int - \csc (θ) \cot (θ) d θ$

Step 2.3.2.2

Multiply the exponents in ${(θ^{2})}^{- 1}$ .

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Step 2.3.2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$y + C_{1} = \int θ^{2 \cdot - 1} d θ + \int - \csc (θ) \cot (θ) d θ$

Step 2.3.2.2.2

Multiply $2$ by $- 1$ .

$y + C_{1} = \int θ^{- 2} d θ + \int - \csc (θ) \cot (θ) d θ$

Step 2.3.3

By the Power Rule, the integral of $θ^{- 2}$ with respect to $θ$ is $- θ^{- 1}$ .

$y + C_{1} = - θ^{- 1} + C_{2} + \int - \csc (θ) \cot (θ) d θ$

Step 2.3.4

Since $- 1$ is constant with respect to $θ$ , move $- 1$ out of the integral.

$y + C_{1} = - θ^{- 1} + C_{2} - \int \csc (θ) \cot (θ) d θ$

Step 2.3.5

Since the derivative of $- \csc (θ)$ is $\csc (θ) \cot (θ)$ , the integral of $\csc (θ) \cot (θ)$ is $- \csc (θ)$ .

$y + C_{1} = - θ^{- 1} + C_{2} - (- \csc (θ) + C_{3})$

Step 2.3.6

Simplify.

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Step 2.3.6.1

Simplify.

$y + C_{1} = - \frac{1}{θ} - - \csc (θ) + C_{4}$

Step 2.3.6.2

Simplify.

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Step 2.3.6.2.1

Multiply $- 1$ by $- 1$ .

$y + C_{1} = - \frac{1}{θ} + 1 \csc (θ) + C_{4}$

Step 2.3.6.2.2

Multiply $\csc (θ)$ by $1$ .

$y + C_{1} = - \frac{1}{θ} + \csc (θ) + C_{4}$

Step 2.4

Group the constant of integration on the right side as $K$ .

$y = - \frac{1}{θ} + \csc (θ) + K$