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 cos**d*, and it makes 16 – *x ^{ }*2 = 16 – 16 sin

Also, *x ^{ }*2 = 16 sin

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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sine substitution | tan substitution | sec substitution | practice | solutions |

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(*memory tweak:*** t*** for ***t***an looks like **+,** use tan substitution with a ^{ 2} *

When the integrand includes the expression ,

we set *x = a tan * which makes *dx = a sec ^{ 2} *

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

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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sine substitution | tan substitution | sec substitution | practice | solutions |

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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

When we factor out 9, replace (*sec ^{ }*2 – 1

With this trig sub, the integral becomes:

Now, replace* tan ^{ 2}*

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

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sine substitution | tan substitution | sec substitution | practice | solutions |

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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sine substitution | tan substitution | sec substitution | practice | solutions |

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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 *

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} *

When we factor out *9*, replace (*1 + **tan ^{ 2} *) with

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} *

and *(36 + x ^{ 2} )^{ 2} = 36^{ 2}*(

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} *

The result is

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sine substitution | tan substitution | sec substitution | practice | solutions |

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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} *

and .

Area = (a boomerang integral)

Solving with integration by parts

u = sec x |
dv = sec ^{ 2 }x dx |
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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 to left side^{ 3} x dx |
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 – 4*x ^{ }*2 = 4(4 –

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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sine substitution | tan substitution | sec substitution | practice | solutions |

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