# 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`.

Answer

`int3\ cos^3x\ dx`

`=3int(cos^2x)cos x\ dx`

`=3int(1-sin^2x)cos x\ dx`

`=3int(cos x\ -sin^2x\ cos x\) dx`

Letting `u=sin x`, and thus `du=cos x\ dx` gives:

`=3[sin x+K_1-intu^2du]`

`=3[sin x-(u^3)/3+K]`

`=3[sin x-(sin^3x)/3+K]`

## 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`.

Answer

`intcos^2 2x\ dx`

Let `u=2x`, then `du=2\ dx`

This gives:

`intcos^2u(du)/2 =1/2intcos^2 u\ du`

`=1/2int(1+cos 2u)/2du`

`=1/4int1+cos 2u\ du`

`=1/4[u+(sin 2u)/2]+K`

`=1/4[2x+(sin 2(2x))/2]+K`

`=x/2+(sin 4x)/8+K`

### Example 3

Integrate: `6intcot^3x\ dx`.

Answer

We have:

`6intcot^3x\ dx =6int(cot^2 x)cot x\ dx`

`=6int(csc^2x-1)cot x\ dx`

`=6(intcsc^2 x\ cot x\ dx` `{:-intcot x\ dx)`

Take the `intcsc^2x\ cot x\ dx` integral first:

Let `u = cot x`, then `du = -csc^2x\ dx`

So

`intcsc^2x\ cot x\ dx =-intu\ du`

`=-u^2/2+C`

`=-(cot^2x)/2+C`

For the second integral, from our table, we have

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

Returning to the main question:

`6intcot^3x\ dx =6(intcsc^2x\ cot x\ dx-` `{:intcot x\ dx)`

`=6(-(cot^2x)/2-ln\ |sin x|)+K`

`=-3\ cot^2x-6\ ln\ |sin x|+K`

## 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.*

(See Period of Sine and Cosine if you are not sure about this.)

### 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.*

Answer

In this case, *T* = 2π.

So

`i_("rms") =sqrt(1/Tint_0^Ti^2dt)`

`=sqrt(1/(2pi)int_0^(2pi)(3+2\ cos t)^2dt)`

Now

`(3+2\ cos t)^2=9+12\ cos t+4\ cos^2 t`

Since `cos 2t = 2 cos^2 t - 1`, it follows that `cos^2t=(cos 2t+1)/2`.

So

`(3+2\ cos t)^2 =9+12 cos t+4((cos 2t+1)/2)`

`=9+12 cos t+2(cos 2t+1)`

`=11+12\ cos t+2\ cos 2t`

So

`int_0^Ti^2dt =int_0^(2pi)(11+12\ cos t+2\ cos 2t)dt`

`=[11t+12\ sin t+sin 2t]_0^(2pi)`

`=22pi`

And, finally:

`i_("rms") =sqrt(1/Tint_0^Ti^2dt) `

`=sqrt((22pi)/(2pi))`

`=sqrt11`

`=3.317`

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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* = *i*_{0} sin ω*t*, show that the root-mean-square of the current for one period
is `(i_0)/sqrt2.`

Answer

In this case,

`T=(2pi)/omega`

So

`i_("rms")=sqrt(1/T int_0^Ti^2dt)`

`=sqrt(omega/(2pi)int_0^((2pi)/omega)(i_0\ sin omegat)^2dt)`

`=sqrt((omega(i _0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)`

Let's just take the integral part first, and put

`sin^2omegat=(1-cos 2 omegat)/2`

`int_0^(2pi//omega)sin^2omegat\ dt =1/2int_0^(2pi//omega)(1-cos 2 omegat)dt`

`=1/2[t-(sin 2omega t)/(2omega)]_0^(2pi//omega)`

`=1/2[((2pi)/omega-0)-(0-0)]`

`=pi/omega`

So

`i_("rms")=sqrt((omega(i_0)^2)/(2pi)int_0^((2pi)/omega)sin^2omegat\ dt)`

`=sqrt{(omega (i_0)^2)/(2pi)pi/omega}`

`=sqrt(((i_0)^2)/2)`

`=(i_0)/sqrt2`

This is what the question required us to show.

Easy to understand math videos:

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## Exercises

Integrate each of the given functions:

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

Answer

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

Write `sin^3x ` `= sin^2x\ sin x` ` = (1 - cos^2x)(sin x)`

So, taking the **indefinite case** first:

`sqrt(cos x) sin^3x`

`=cos^(1//2)x(1-cos^2x)(sin x)`

`=(cos^(1//2)x-cos^(5//2)x)(sin x)`

Put `u = cos x` then `du = -sin x\ dx`

So

`intsqrt(cos x)\ sin^3x\ dx =int(cos^(1//2)x-cos^(5//2)x)(sin x)dx`

`=-int(u^(1//2)-u^(5//2))du`

`=-(2u^(3//2))/3+(2u^(7//2))/7+K`

`=-(2\ cos^(3//2) x)/3+(2\ cos^(7//2)x)/7+K`

So we have for the **definite case**:

`int_(pi//3)^(pi//2)sqrt(cos x)\ sin^3x\ dx =[-(2 cos^(3//2)x)/3+(2 cos^(7//2)x)/7]_(pi//3)^(pi//2)`

`=[(0+0)-(-0.23570+0.02525)]`

`=0.2104`

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`

Answer

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

We recall that

`2\ sin^2theta=1-cos 2theta`

In this case, if `θ = 4x`,

`2 sin^2 4x` `=1-cos [2(4x)]` `=1-cos 8x`

So `sin^2 4x=(1-cos 8x)/2`

`int_0^1sin^2 4x dx =1/2int_0^1(1-cos 8x) dx`

`=1/2[x-(sin 8x)/8]_0^1`

`=1/2[(1-0.9894/8)-(0-0)]`

`=0.4382`

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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`

Answer

`int cot 4x\ csc^4 4x\ dx`

We write the expression under the integral sign as follows:

`cot 4x\ csc^4 4x\ dx` `=(csc^3 4x)\ cot 4x\ csc\ 4x`

Then, let `u = csc\ 4x` and so we have `du = -4\ csc\ 4x\ cot 4x\ dx`

That is, `-(du)/4 = csc\ 4x\ cot 4x\ dx`

Now we can perform the integral:

`int cot 4x\ csc^4 4x\ dx =int(csc^3 4x)\ cot 4x\ csc\ 4x\ dx`

`=-1/4intu^3du`

`=-u^4/16+K`

`=-(csc^4 4x)/16+K`

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

Answer

Recall that

`sec^2x=1+tan^2x`

So we can write the part under the integral as:

`sqrt(tan x)\ sec^4 x =(tan^(1//2) x)\ sec^2 x\ sec^2 x`

`=(tan^(1//2) x )(1+tan^2 x )sec^2 x`

`=(tan^(1//2) x+tan^(5//2)x)sec^2 x`

Next, we put `u = tan x`, giving `du = sec^2x\ dx`

So our integral becomes:

`int (tan^(1//2) x+tan^(5//2)x)sec^2 x\ dx =int(u^(1//2)+u^(5//2))du`

`=2/3u^(3//2)+2/7u^(7//2)+K`

`=2/3tan^(3//2)x+2/7tan^(7//2)x+K`

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

Answer

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

We need to re-express this and we make use of a fairly common technique. We multiply the numerator (top) and denominator (bottom) of the fraction by the **conjugate** of the denominator. (The conjugate has the opposite sign in the middle. In this case, it's a minus sign.)

We will make use of this important result:

`sin^2 x+ cos^2 x = 1`

`1/(1+sin x) =1/(1+sin x)xx(1-sin x)/(1-sin x)`

`=(1-sin x)/(1-sin^2x)`

`=(1-sin x)/(cos^2 x)`

`=sec^2x-(sin x)/(cos^2 x)`

Now we can integrate.

Putting `u = cos x` in the right hand part, we have:

`du = -sin x\ dx`

So

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

`=2int_(pi//6)^(pi//3)(sec^2x-(sin x)/(cos^2x))dx`

`=2[tan x-1/(cos x)]_(pi//6)^(pi//3)`

`=2[(1.73215-2)-` `{:(0.57735-1.15470)]`

`=0.6190`

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`.

Answer

The curve in this problem is `y = ln cos x`.

We need its derivative:

`(dy)/(dx)=-(sin x)/(cos x)=-tan x`

For this problem, we'll make use of an earlier result,

`1+tan^2 x=sec^2 x`

Applying the formula gives:

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

`=int_0^(pi//3)sqrt(1+(-tan x)^2)dx`

`=int_0^(pi//3)sqrt(1+tan^2x) dx`

`=int_0^(pi//3)sqrt(sec^2x) dx`

`=int_0^(pi//3)sec x\ dx`

`=[ln\ |sec x+tan x|]_0^(pi//3)`

`=[ln\ |sec (pi/3)+tan (pi/3)|` `{:-|ln(1)+ln(0)|]`

`=1.317`

Here's the graph of the arc length we just found (in pink). 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 (pink).

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