# Trigonometric Substitution: Solved Integral Calculus Problems

TR Smith is a product designer and former teacher who uses math in her work every day.

Trigonometric substitution is a clever integration technique used to find the antiderivatives of functions that involve the square root of a quadratic polynomial, i.e., a second degree polynomial of the form Px^2 + Qx + R. There are three substitution rules that will turn such integrands into trigonometric functions, which are easier to work out.

• If the integral contains sqrt(x^2 - a^2) use x = a*sec(u) and dx = a*sec(u)*tan(u) du.
• If the integral contains sqrt(x^2 + a^2) use x = a*tan(u) and dx = a*sec(u)^2 du.
• If the integral contains sqrt(a^2 - x^2) use x = a*sin(u) and dx = a*cos(u) du.
• Alternatively, if the integral contains sqrt(a^2 - x^2) use x = a*cos(u) and dx = -a*sin(u) du.

With these three substitution rules you may also need to know some trigonometric identities:

• sin(u)^2 + cos(u)^2 = 1
• tan(u)^2 + 1 = sec(u)^2
• tan(arccos(u)) = sqrt(1 - u^2)/u
• tan(arcsin(u)) = u/sqrt(1 - u^2)

A general second degree polynomial can be converted to one of the three forms via the method of "completing the square." A quick review of how to complete the square is shown at the end of this tutorial. Here are several example problems for each of the three cases, so you can see how each rule is applied and how they make solving square root-quadratic integrals much easier.

## Example 1: Integral of sqrt(x^2 + 4x + 29)

To integrate the function f(x) = sqrt(x^2 + 4x + 29), we first need to complete the square and rewrite it as

∫ sqrt(x^2 + 4x + 29) dx = ∫ sqrt[(x+2)^2 + 5^2] dx

Now the function is essentially in the form sqrt(x^2 + a^2) and we can use the second rule of trigonometric substitution to find the antiderivative. We make the change of variables

x + 2 = 5*tan(u) and dx = 5*sec(u)^2 du

If we plug this variable substitution into the integral, we get

∫ sqrt[(x+2)^2 + 5^2] dx = ∫ sqrt[(5*tan(u))^2 + 5^2] * 5*sec(u)^2 du

= ∫ 25*sec(u)^3 du

The integral of secant cubed was worked out in a previous article. Using the antiderivative formula from that calculus tutorial, we now have

∫ 25*sec(u)^3 du = (25/2)sec(u)tan(u) + (25/2)*LN( sec(u) + tan(u) ) + C

Our original variable is x, so now we must convert this expression into a function of x. Since we have x + 2 = 5*sec(u), we have sec(u) = (x+2)/5, and tan(u) = (1/5)sqrt(x^2 + 4x + 29). Putting everything together, we get the final answer for the antiderivative of the original function.

∫ sqrt(x^2 + 4x + 29) dx

= (1/2)(x+2)*sqrt(x^2 + 4x + 29)
+ (25/2)*LN( (x+2)/5 + (1/5)sqrt(x^2 + 4x + 29) ) + C

Since C is an unspecified constant of integration, it can be replaced with C + LN(5), which allows you to express the integral in an equivalent, but cleaner form:

∫ sqrt(x^2 + 4x + 29) dx

= (1/2)(x+2)*sqrt(x^2 + 4x + 29) + (25/2)*LN( x + 2 + sqrt(x^2 + 4x + 29) ) + C

## Example 2: Integral of 1/[x*sqrt(4 - x^2)]

To find the antiderivative of f(x) = 1/[x*sqrt(4 - x^2)], we use the third trigonometric substitution rule with

x = 2*sin(u) and dx = 2*cos(u) du

This change of variables gives us the new integral

∫ 1/[2*sin(u) * 2*cos(u)] * 2*cos(u) du

= (1/2)*∫ 1/sin(u) du

= -(1/2)*LN( cot(u) + csc(u) ) + C

Now we need to replace the variable u with a function of x. Since x = 2*sin(u), we have u = arcsin(x/2). Plugging this into the expression above and using properties of inverse trig functions gives us

-(1/2)*LN( cot(arcsin(x/2)) + csc(arcsin(x/2)) ) + C

= -(1/2)*LN( sqrt(4 - x^2)/x + 2/x ) + C

The expression in bold above is the antiderivative of the function we started with. We can put it into a more elegant form by using properties of logarithms. This gives us the final answer

∫ 1/[x*sqrt(4 - x^2)] dx = 0.5*LN(x) - 0.5*LN( sqrt(4 - x^2) + 2) + C

## Example 3: Integral of sqrt(x^2 - 16)/x

This integral calls for the first trigonometric substitution rule, x = 4*sec(u). Taking the derivative of both sides of the equation x = 4*sec(u) gives us the differentials dx = 4*sec(u)tan(u) du. Now we can make the replacements in the original integral.

∫ sqrt(x^2 - 16)/x dx

= ∫ [sqrt(16*sec(u)^2 - 16)/(4*sec(u))] * 4*sec(u)tan(u) du

= ∫ 4*tan(u)^2 du

= ∫ 4(sec(u)^2 - 1) du

= ∫ 4*sec(u)^2 du - ∫ 4 du

= 4*tan(u) - 4u + C

Now we make the replacement for the variable u as a function of x. To do this, we must solve the equation x = 4*sec(u) to isolate u, which gives us u = arccos(4/x). Plugging this into the expression immediately above gives us

4*tan(u) - 4u + C

= 4*tan(arccos(4/x)) - 4*arccos(4/x) + C

Although the antiderivative form above is correct, it is usually presented in a more standard form. To convert it to an equivalent form, we use the identities

arccos(4/x) = π/2 - arcsin(4/x) = -arctan(4/sqrt(x^2-4))

tan(arccos(4/x)) = (1/4)*sqrt(x^2 - 16)

This gives us

4*tan(arccos(4/x)) - 4*arccos(4/x) + C

= sqrt(x^2 - 16) + 4*arcsin(4/x) + π/2 + C

= sqrt(x^2 - 16) + 4*arcsin(4/x) + C

= sqrt(x^2 - 16) + 4*arctan(4 / sqrt(x^2 - 16)) + C

since the addition of π/2 to the arbitrary constant simply yields another arbitrary constant.

## How to Complete the Square

To simplify sqrt(Px^2 + Qx + R) and put it into a usable form for trig substitution we apply a little algebra to the radical. If P is a positive constant, we apply the steps

sqrt( Px^2 + Qx + R)

= sqrt(P) * sqrt( x^2 + (Q/P)x + R/P )

= sqrt(P) * sqrt[ (x + Q/(2P))^2 + R/P - Q^2/(4P^2) ]

Here, the element (x + Q/(2P))^2 acts like "x^2" and the element R/P - Q^2/(4P^2) acts like "a^2" from the elementary forms "x^2 + a^2" and "x^2 - a^2."

If P is negative, then we apply the same process, but instead of the factor sqrt(P) we use sqrt(-P). This gives us

sqrt( Px^2 + Qx + R )

= sqrt(-P) * sqrt[ -x^2 - (Q/P)x - R/P ]

= sqrt(-P) * sqrt[ Q^2/(4P^2) - R/P - (x + Q/(2P))^2 ]

Here, the element -(x + Q/(2P))^2 acts like "-x^2" and the element Q^2/(4P^2) - R/P acts like "a^2" from then elementary form "a^2 - x^2." In order for the antiderivative to exist, Q^2/(4P^2) - R/P must be non-negative.

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