Integral of ∫sec^2(x)dx

The integral of sec²(x) should be a very straight forward operation, because early in our calculus studies, we would have learned that the derivative of tan(x) is equal to sec²(x). That is...

\frac{\mathrm{d} }{\mathrm{d} x}\tan x=\sec^2x

And since integration is the reverse of differentiation, it follows then that...

\int \sec^2x=\tan x+C

However, what if you didn't know this result in the first place?

Well, ironically, you would need to know quite a higher level of trigonometric identities and calculus to complete this integral.

In this video, we explore 2 ways of performing this integral:

1. Integration by parts
2. Using the t-substitution and partial fraction decomposition.

Method 1:

With method 1, we use the Pythagorean Identity: sec²(x) = tan²(x) + 1, so we can write...

\tan^2 x=\frac{\sin^2x}{\cos^2x}=-\sin x\cdot \frac{-\sin x}{\cos^2x}

And thus we have the integration of 2 parts: u = -sin(x) and dv = -sinx/cos²x

Method 2:

With method 2, we use the substitution t = tan(x/2), and for this, we also need the relationship between cosx and tan(x/2), which is...

\cos x=\frac{1-t^2}{1+t^2}

And we'll also need...

\mathrm{d} x=\frac{2}{1+t^2}\mathrm{d} t

With the expression for cosx, we can express...

\sec x=\frac{1+t^2}{1-t^2}

And thus...

\sec^2 x=\frac{\left (1+t^2 \right )^2}{\left (1-t^2 \right )^2}

The integral of sec^2(x) then becomes...

\int \sec^2x\mathrm{d}x=\int \frac{2\left ( 1+t^2 \right )}{\left (1-t^2 \right )^2}\mathrm{d}t