# 5. Integration: Other Trigonometric Forms

by M. Bourne

We can use the trigonometric identities that we learned earlier to simplify the integration process.

The main identities are shown here for reference:

cos^2x+sin^2x=1

1+tan^2x=sec^2x

1+cot^2x=csc^2x

2\ cos^2x=1+cos\ 2x

2\ sin^2x=1-cos\ 2x

The process that we use involves using the trigonometric ratios to simplify the expression, or to get the expression into a form that can be integrated.

## Integrating a Product of Powers of Sine and Cosine - one power odd

To integrate a product of powers of sine and cosine, we use

cos^2x+sin^2x=1

if at least one of the powers is odd.

### Example 1

Integrate: int3\ cos^3x\ dx.

## Integrating a Product of Powers of Sine and Cosine - powers even

We use

2\ cos^2x=1+cos\ 2x

or

2\ sin^2x=1-cos\ 2x

if the power of sin\ x or cos\ x is even.

### Example 2

Integrate: intcos^2\ 2x\ dx.

### Example 3

Integrate: 6intcot^3x\ dx.

## Application - Root Mean Square Value

The root mean square value of the function y with respect to x is given by:

y_("rms")=sqrt(1/T int_0^T y^2dx

where T is the period of y.

### Effective Current

A common use of this concept is effective current. This is the value of the direct current that would produce the same quantity of heat energy in the same time as a certain alternating current. It is used in the design of heaters.

### Example 4

Find the root mean square (rms) value of i = 3 + 2 cos t.

This is a graph of our cosine current with the RMS effective current shown.

Graph of i(t)=3+2cos(t), with the RMS current indicated by the dashed magenta line.

### Example 5

For a current i given by i = i0 sin ωt, show that the root-mean-square of the current for one period is (i_0)/sqrt2.

## Exercises

Integrate each of the given functions:

1. int_(pi//3)^(pi//2)sqrt(cos\ x)\ sin^3x\ dx

The solution for Exercise 1 represents the area under the curve y(x)=sqrt(cos\ x)\ sin^3x\ dx between pi/3 <= x <= pi/2. Here is that graph:

Graph of y(x)=sqrt(cos x)\ sin^3x, indicating the area under the curve from x=pi/3 to x=pi/2.

Zooming out that graph shows it's periodic (it repeats itself) with period 2pi. There are holes in the graph because of the sqrt(cos x) part (we can't have the square root of a negative number).

Graph of y(x)=sqrt(cos x)\ sin^3x dx, zoomed out to see its periodic nature.

2. int_0^1sin^2 4x\ dx

Here is the graph of the integration we just found, indicating the area under the curve y=sin^2 4x.

Graph of y(x)=sin^2 4x, indicating the area under the curve from x=0 to x=1.

3. intcot\ 4x\ csc^4 4x\ dx

4. intsqrt(tan\ x)\ sec^4x\ dx

5. int_(pi//6)^(pi//3)(2dx)/(1+sin\ x

Graph of y(x)=2/(1+sin(x)), indicating the area under the curve from x=pi/6 to x=pi/3.

### Application - Length of a Curve

The length s of the arc of a curve y = f(x) from x = a to x = b is given by:

s=int_a^bsqrt(1+((dy)/dx)^2dx

Find the length of the curve y = ln (cos x) from x=0 to x=pi/3.

Here's the graph of the arc length we just found. I needed to take the absolute value of the cos(x) values, otherwise there would be gaps in the graph (whenever cos(x) was negative).

Graph of y(x)=ln|cos x|, with the curve length we just found indicated in magenta.

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