Simplifying Integrals with Trigonometric Substitutions (Part I)

Trigonometric Substitutions (Part I)

There is a special kind of integration called trigonometric substitution, which is useful for handling certain kinds of integrals. The feature of this integration is that we have a square root of some quadratic expression here.

∫dxa2+x2. The key idea is to simplify, with suitable substitution a2+ x

The general form of an integral with a square root is difficult to simplify. The key idea for simplifying this integral is to make a substitution for x so that we can remove the square root.

Once we can remove the square root, things will become simpler. The expression inside here is a square plus x squared.

x = θ, −π2 < θ <π2

Suppose we set x = a tan(theta), where theta is defined as being between -pi/2 and pi/2.

a2+ x2 = a21 + tan2( θ) = a2sec2θ

Then we realize that a squared plus x squared becomes a square plus tan(theta). We use the trigonometric identity that one plus tangent theta squared equals secant theta squared.

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So, this substitution allows us to turn z squared plus x squared into a square.

a2+ x2 = a|sec θ | = a sec θ, since sec θ > 0 for −π2 < θ <π2

When we take the square root of a squared plus x square, we just take the square root here and we get an absolute value of secant theta. Remember, when we take square root, we always end up with something non-negative. That's why we have to add the absolute value sign.

This is also inconvenient. However, in the range of theta that we have chosen in this range between minus pi over two and pi over two, secant theta is greater than or equal to zero; therefore it is unnecessary to include an absolute value sign.

dx = ad(tan θ) = a sec2θ dθ

The integral can be simplified into the following form:

∫dxa2+x2 = ∫1a sec θa sec2θ dθ = ∫ sec θ dθ

We are integrating this expression, and the denominator here is equal to a times secant theta. So what about dx? x is equal to that. When we differentiate, dx equals a differentiate tangent theta which is secant theta squared times d theta. The cancellation can take place, and what ends up is this expression which is well known: it's logarithm of absolute value of secant theta plus tangent theta plus C.

∫dxa2+x2 = ∫1a sec θa sec2θ dθ = ∫ sec θ dθ

= ln |sec θ + tan θ | + C

We now have to convert back to x because the original integral was in terms of x.

Let us consider a right-angled triangle in which the vertical side is x, the horizontal side is a, and the two sides are related by x over a = tangent(theta). This means that this angle is equal to theta. We then use the Pythagorean theorem to see that the limits of this hypotenuse are given by sqrt(a^2) + x^2. From here we clearly see that sec(theta) = 1/a and tan(theta) = x/a.

∫dxa2+x2 = ∫1a sec θa sec2θ dθ = ∫ sec θ dθ

= ln |sec θ + tan θ | + C = ln a2+x2a+xa||||||+ C

Therefore, we can convert this expression back in terms of x. In the previous example, we saw how to make use of a trigonometric substitution to simplify √(x)2 + x2.

Example: ∫x2a2−x2dx

Now let's look at another type of integral, in which we have to simplify something similar. In this example, the most difficult thing is the square root of a square minus x square. So we try to find a substitution of x in terms of a trigonometric function that would allow us to simplify this one or in other words make a square minus x square into a square.

a^2-x^2

Given this situation, we can set x equal to a sin(theta) where theta is between -pi/2 and pi/2.

x = a sin θ, −π2 < θ <π2

a2 − x2 = a2(1 − sin2θ) = a2cos2θ

Substituting x with the expression in parentheses, we see that a square minus x square equals a square minus a square sin theta square. Recall that one minus sin theta square is cosine theta square, so this becomes a square now. Taking square root, we end up with this one, with the absolute value sign.

2 − x2 = a2(1 − sin2θ) = a2cos2θ

= a|cos θ | = a cos θ for −π2 < θ <π2

In the range of values for theta shown here, we see that cosine theta is always greater than or equal to zero. In other words, again this absolute value sign is unnecessary and we end up with this result.

Example: ∫x2a2−x2dx

Since dx = a cos θ dθ

As before, we find from this substitution that dx is equal to a derivative of sin theta, which is cosine theta times d theta.

∫x2a2−x2dx = ∫a2sin2θa |cos θ |a cos θ dθ

We find the integral of the sine function squared by adding together the squares of the cosine and x. By integrating these two terms, we arrive at the formula shown above.