# How to Integrate Sec(x)^3 and Sec(x)Tan(x)^2 -- Trig Antiderivatives

Finding the antiderivatives of f(x) = sec(x)^3 and g(x) = tan(x)^2*sec(x) are challenging problems in integral calculus because standard tricks like integration by parts, substitution, and using the identity sec(x)^2 = tan(x)^2 + 1 *seemingly* don't work to reduce the integrals into simpler, easier forms.

What you find when you try to integrate either function with standard integration techniques is another integral expression involving the other function! However, what looks like a process of going in circles is precisely the tool you can use to find their antiderivatives. The clever trick used to find the integrals of sec(x)^3 and sec(x)tan(x)^2 is similar that of integrating the functions sin(x)e^x and cos(x)e^x.

## Antiderivative of Sec(x)^3

First we apply the trigonometric identity sec(x)^2 = tan(x)^2 + 1 to obtain

∫ sec(x)^3 dx

= ∫ sec(x)[tan(x)^2 + 1] dx

= ∫ sec(x)tan(x)^2 dx + ∫ sec(x) dx

= ∫ sec(x)tan(x)^2 dx + LN | sec(x) + tan(x) | + C

This procedure gives us an integral expression for sec(x)^3 in terms of the integral sec(x)tan(x)^2. Let's try the technique of integration by parts on the integral of sec(x)tan(x)^2 using, using the assignment

u = tan(x) and

dv = sec(x)tan(x) dx

This also gives us

du = sec(x)^2 dx and

v = sec(x)

Now we have

∫ sec(x)tan(x)^2 dx = uv - ∫ v*du

= sec(x)tan(x) - ∫ sec(x)sec(x)^2 dx

= sec(x)tan(x) - ∫ sec(x)^3 dx

And now we are back with another integral of sec(x)^3. But this work was not all for naught! If we combine this with the decomposition we did at the beginning, we see that

∫ sec(x)^3 dx

= ∫ sec(x)tan(x)^2 dx + LN | sec(x) + tan(x) | + C

= sec(x)tan(x) - ∫ sec(x)^3 dx + LN | sec(x) + tan(x) | + C

Now we have an equation with the integral of secant cubed on both sides of the equation. If we add the integral to both sides, it cancels out on one side and gets doubled on the other:

2 * ∫ sec(x)^3 dx = sec(x)tan(x) + LN | sec(x) + tan(x) | + C

Finally, we can divide both sides by 2 to obtain the antiderivative of sec(x)^3:

**∫ sec(x)^3 dx = 0.5*sec(x)tan(x) + 0.5*LN | sec(x) + tan(x) | + C**

## Antiderivative of Sec(x)Tan(x)^2

From the previous section we found that

∫ sec(x)tan(x)^2 dx = sec(x)tan(x) - ∫ secc(x)^3 dx

But since we now know the integral of secant cubed, we can rewrite this as

∫ sec(x)tan(x)^2 dx

= sec(x)tan(x) - 0.5*sec(x)tan(x) - 0.5*LN | sec(x) + tan(x) | + C**= 0.5*sec(x)tan(x) - 0.5*LN | sec(x) + tan(x) | + C**

## Application of Integration

Knowing the derivatives of trig functions is useful when you apply trig substitution to find the antiderivatives of functions involving sqrt(1 - x^2) and similar functions. For example, let's find the antiderivative of the function h(x) = sqr(1 - x^2)/x^3 using the substitution

x = cos(u)

arccos(x) = u

dx = -sin(u) du

This gives us the new integral

∫ sqrt(1 - x^2)/x^3 dx

= ∫ [sqrt(1 - cos(u)^2) / cos(u)^3] * -sin(u) du

= ∫ -sin(u)^2 / cos(u)^3 du

= - ∫ tan(u)^2 * sec(u) du

Since we now know the antiderivative of tan(u)^2*sec(u), we have

-∫ tan(u)^2 * sec(u) dx

= 0.5*sec(u)tan(u) - 0.5*LN | sec(u) + tan(u) | + C

At this point we make the reverse substitution with u = arccos(x), which gives us the antiderivative

**0.5*(1/x)*tan(arccos(x)) - 0.5*LN | 1/x + tan(arccos(x)) | + C**

This expression can be put into an equivalent form using inverse trigonometric identities and properties of logarithms:

**sqrt(1 - x^2)/(2x^2) - 0.5*LN | (1 + sqrt(1 - x^2))/x | + C**

## Comments

Well I have no idea what you said, lol! but this will be so useful for anybody that is studying math, so I voted it useful, nice one, wish I understood...lol!

Nell Rose, I am in the same boat. I admire calculus-geometry for being so brilliant at math yet also writing so well. Not many people have both those abilities.

Thank you for this explanation of how to integrate sec(x)^3! I'm taking Calculus II with Analytic Geometry and it can be rather overwhelming. We have to know how to integrate sec(x)^3 and my textbook doesn't adequately explain it.

5