# 4. Integration: Basic Trigonometric Forms

by M. Bourne

We obtain the following integral formulas by reversing the formulas for differentiation of trigonometric functions that we met earlier:

`int sin\ u\ du=-cos\ u+K`

`int cos\ u\ du=sin\ u+K`

`int sec^2u\ du=tan\ u+K`

`int csc^2u\ du=-cot\ u+K`

We now apply the power formula to integrate some examples.

**NOTE:** All angles in this section are in **radians**. The formulas don't work in degrees.

**Example 1: **Integrate:** **`inte^xcsc^2(e^x)dx`

**Example 2: **Integrate:** **`int(sin(1/x))/(x^2)dx`

## Integral of sec *x*, csc *x*

These are obtained by simply reversing the differentiation process.

`int sec\ u\ tan\ u\ du=sec\ u+K`

`int csc\ u\ cot\ u\ du=-csc\ u+K`

**Example 3:** Integrate:** **`int csc\ 2x\ cot\ 2x\ dx`

## Integral of tan *x*, cot *x*

Now, if we want to find `int tan x\ dx`, we note that

`int tan x\ dx=int(sin x)/(cos x)dx`

Let `u=cos x`, then `du=-sin x\ dx`. Our integral becomes:

`int tan x\ dx=int(sin x)/(cos x)dx`

`=-int(du)/u`

`=-ln |u|+K`

`=-ln |cos x|+K`

Similarly, it can be shown that

`intcot\ x\ dx=ln\ |sin\ x|+K`

## Summary of Integrals of Trigonometric Functions

We summarise the trigonometric integrals as follows:

`inttan\ u\ du=-ln\ |cos\ u|+K`

`intcot\ u\ du=ln\ |\sin\ u|+K`

`intsec\ u\ du=ln\ |\sec\ u+tan\ u|+K`

`intcsc\ u\ du=ln\ |csc\ u-cot\ u|+K`

**Example 4: **Integrate:** **`intx^2cot\ x^3dx`

**Example 5: **Integrate: `6int_0^1 tan{:x/2:}dx`

**Example 6: **Find the area under the curve of `y = sin\ x` from
`x = 0` to `x=(3pi)/2`**.**

### Exercises

Integrate each of the given functions:

**1.** `int(sin\ 2x)/(cos^2x)dx`

**2.** `int_(pi//4)^(pi//3)(1+sec\ x)^2dx`

**3. **If the current in a
certain electric circuit is *i* = 110 cos 377*t,* find
the expression for the voltage across a 500-μF
capacitor as a function of time. The initial voltage is zero.
Show that the voltage across the capacitor is
90° out of phase with the current.

We need the following result from electronics, which gives the voltage across a capacitor, where *C* is the capacitance:

`V_C=1/Cinti\ dt`

**4.** A force is given as a function of
the distance from the origin as

`F=(2+tan\ x)/(cos\ x`

Express the work done by this force as a function of
*x* if *W* = 0 for *x* =
0.

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