Integrals Solved with Trig Substitutions

There are 3 integrand expressions which, though they often can be solved with an algebraic substitution, sometimes they require a trigonometric substitution.

The 3 expressions are: (a is a real number or constant)

 1) 2) 3)

The substitutions we use are based on 2 trig identities, namely:

1 – sin 2 x = cos 2 x

1 + tan 2 x = sec 2 x

Just as with algebraic substitutions, we will have to unsubstitute once we've integrated the new expression. Let's do a little review of trig function properties.

If we set , we're saying that angle has a sine = x / a. This says that the side opposite to angle = x and the hypotenuse = a if is constructed in a right triangle.

In other words, angle = arcsin (x / a)

In order to unsubstitute, we make a "picture" of angle like this.

Now, if we need an expression for any one of angle 's trig functions, we can figure it out from the "picture". Notice that the 3rd side of the triangle is the ugly expression we eliminated from the integrand with the trig sub. This will ALWAYS happen. The missing side of the triangle depicting angle will always be the same as the mess for which we substituted.

We must make sure however to label the correct sides with the correct variable. Recall that the sine of an angle = opposite / hypotenuse, cosine = adjacent / hypotenuse and
tangent = opposite / adjacent. To get the reciprocal function values, flip the fractions.

Here's how to use each one of these substitutions.

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When the integrand includes the expression ,

we set x = a sin which makes dx = a cos d

and it makes

or simply a cos.

Example 1:

We let x = 4 sin (since a 2 = 16)

This makes dx = 4 cosd, and it makes 16 – x 2 = 16 – 16 sin 2

Also, x 2 = 16 sin 2

Now the integral becomes

Now we look at our image of .

So, .

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Note: When doing definite integrals with trig subs, we don't attempt to change the limits of integration with the changes in variable. It gets too confusing. Once we've integrated and unsubstituted, then we apply the limits to find the numerical value of the integral.

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(memory tweak: t for tan looks like +, use tan substitution with a 2 + x 2 .)

When the integrand includes the expression ,

we set x = a tan which makes dx = a sec 2 d

and it makes

or simply a sec .

Example 2:

We let x = 2 tan (since a 2 = 4)

This makes dx = 2 sec 2 d, and it makes 4 + x 2 = 4 + 4 tan 2 .

When we factor out 4, replace (1 + tan 2 ) with sec 2,

Now,

Now we look at our image of

Returning to the original variable x we get

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When the integrand includes the expression ,

we set x = a sec , which makes dx = a sec tan d

and it makes

or simply a tan .

Example 3:

We let x = 3 sec (since a 2 = 9)

This makes dx = 3 sec tan d, and it makes x 2 – 9 = 9 sec 2 – 9

When we factor out 9, replace (sec 2 – 1 ) with tan 2 so, ,.

With this trig sub, the integral becomes:

Now, replace tan 2 with sec 2 – 1 to get:

Now we look at our image of

Returning to the original variable x we get

Special Note: when the x 2 term in the integrand expression has a coefficient other than 1, factor the constant like this:

1) Use a trigonometric substitution to find:

 a) b) c) d) e) f)

2) Find the area enclosed by the graphs of

 , y = 0, x = 0, x = 3

3) Find the area of the region enclosed by the ellipse 4x 2 + y 2 = 16.

(Hint: put the equation in standard form, draw a diagram, then solve the equation for y = f(x))

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1) a)

We let x = 2 sin (since a 2 = 4)

So dx = 2 cos d, and it makes 4 x 2 = 4 4 sin 2

Also, x 2 = 4 sin 2

Now the integral becomes

Now we use the trig identity 1 + cot 2 x = csc 2 x to get:

Now we look at our image of

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b)

We let x = 3 tan (since a 2 = 9)

so, dx = 3 sec 2 d, and 9 + x 2 = 9 + 9 tan 2

When we factor out 9, replace (1 + tan 2 ) with sec 2 ,

From a table of integrals, we get 1/3 ln | csc – cot |

Now we look at our image of

Returning to the original variable x we get

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c)

Using x = 5 sec , the integral becomes ,

Now we look at our image of

Returning to the original variable x we get

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d)

Substitute x = 3/2 sin (sin = 2x / 3 ) to get

, use the half-angle identity cos 2 A = ½(1 + cos 2A),

the integral becomes

Antidifferentiate and use double-angle identity sin 2A = 2 sin A cos A

then unsubstitute the result:

We get

e)

Let x = 6 tan , so dx = 6 sec 2 d ,

and (36 + x 2 ) 2 = 36 2(1 + tan 2 ) 2 = 36 2(sec 4 )

After substitution we get

which we solved in the previous question.

Be careful however, since in this question, we defined h using the tangent function, not sine. So here, = arctan (x / 6 )

The solution is

f)

After substitution we get , which is an odd power of cosine.

Break it up into cos 2 cos d , apply the identity cos 2 A = 1 – sin 2 A, then integrate.

The result is

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2) Find the area enclosed by the graphs of

 , y = 0, x = 0, x = 3

Area =

Substitute x = 2 tan , therefore dx = 2 sec 2 d ,

and .

Area = (a boomerang integral)

Solving with integration by parts

 u = sec x dv = sec 2 x dx . du = sec x tan x dx v = tan x sec 3 x dx = sec x tan x – sec x tan 2 x dx sec 3 x dx = sec x tan x – sec x (sec 2 x – 1) dx sec 3 x dx = sec x tan x – sec 3 x dx + sec x dx transpose – ò sec 3 x dx to left side 2sec 3 x dx = sec x tan x + sec x dx integrate ò sec x dx, divide by 2 sec 3 x dx = ½ ( sec x tan x + ln | sec x + tan x | ) + C

Since we have 4 times this integral we need to evaluate

2 ( sec tan + ln | sec + tan | )

Using the image of from our substitution of x / 2 = tan

we get

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3) Find the area of the region enclosed by the ellipse 4x 2 + y 2 = 16.

When we divide by 16 we get . Let's draw it.

Now we solve for y = f (x)

Since y 2 = 16 – 4x 2 = 4(4 – x 2 ),
Since the ellipse is symmetric to both axes, the Area we want
can be found by taking 4 times the area in the first quadrant.

, when we substitute x = 2 sin
Now we apply the half-angle formula to get 16
Now antidifferentiate, and apply the double-angle formula for sin
From the diagram of the integral becomes

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