Integral of sec²(x)|Two ways

28 Jun 2020 Leave a Comment

Hi everyone! I know it’s been a really long time since my last post, but lessons during COVID-19 really had me busy and stressed out… but here I am again! Hope you are all doing great and ready for more integrals!

Graph of the function f(x)=sec²(x) in red and the F(x) in blue.

In this post I am going to show you two methods I found to integrate the function y=sec²x. The first one requires a bit of knowledge about trigonometry and trigonometric identities and might not be very intuitive, whereas the second involves a substitution right at the beginning and then some algebra and trigonometric identities. Feel free to share more methods if you know them!

First method

$\int \sec^2(x)\,\mathrm{d}x$

First of all, what is sec(x) equal to? It’s simply the reciprocal of the function cos(x), i.e. 1/cos(x). Therefore, secant squared is equal to one over cosine squared. That simple!

$\int \frac{1}{\cos^2(x)}\,\mathrm{d}x$

Now comes the not very intuitive part. In trigonometry, 1 is given by the sum of the squares of the two fundamental functions sine and cosine. This is the most important trigonometric identity and the one we are going to use:

$\int\frac{\sin^2(x)+\cos^2(x)}{\cos^2(x)}\,\mathrm{d}x$

At this point we basically have

$\int\left [\frac{\sin^2(x)}{\cos^2(x)}+\frac{\cos^2(x)}{\cos^2(x)} \right ]\mathrm{d}x$

It’s like (a+b)*(1/c)=a/c+b/c: we are simply distributing the 1/c.

$\int\left [\frac{\sin^2(x)}{\cos^2(x)}+1 \right ]\mathrm{d}x$

Let’s apply linearity:

$\int\frac{\sin^2(x)}{\cos^2(x)}\,\mathrm{d}x +\int 1\, \mathrm{d}x$

Remember the property

$\int m\,\mathrm{d}x=mx+C$

where m is a constant, in our case 1; therefore

$x+\int\frac{\sin^2(x)}{\cos^2(x)}\,\mathrm{d}x$

The problem here is mostly the fact that we have sine squared at the numerator. If it was simply sin(x), we would have performed a u-substitution by letting u=cos(x) and its derivative, -sin(x), would have gone to the denominator and so it would have cancelled out with the sin(x) on top.

$\int\frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x$ $u=\cos(x)\;\;\;\;\;\mathrm{d}u=-\sin(x)\mathrm{d}x\rightarrow \mathrm{d}x=-\frac{\mathrm{d}u}{\sin(x)}$ $\int \frac{{\color{Red} \sin(x)}}{\cos^2(x)}\left ( -\frac{1}{{\color{Red} \sin(x)}} \right )\mathrm{d}u$ $\int \frac{{\color{Red} \sin(x)}}{u^2}\left ( -\frac{1}{{\color{Red} \sin(x)}} \right )\mathrm{d}u$

Notice that cos²(x) becomes u² because we let u=cos(x).

$-\int \frac{1}{u^2}\,\mathrm{d}u=-\int u^{-2}\,\mathrm{d}u$

Let’s apply the power rule for integrals. Just in case you don’t remember,

$\int x^n\,\mathrm{d}x=\frac{x^{n+1}}{n+1}+C$ $- \frac{u^{-1}}{-1}\,\mathrm{d}u=u^{-1}=\frac{1}{u}$

Let’s undo the substitution (remember: u=cos(x))

$\int \frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x=\frac{1}{\cos(x)}+C$

BUT we don’t have this integral; instead we have

$x+\int \frac{\sin^2(x)}{\cos^2(x)}\,\mathrm{d}x$

But after all, sin²(x) is sin(x)*sin(x), so:

$\int \sin(x)\cdot \frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x$

We have a product, which means we can perform integration by parts, and naturally we are going to differentiate sin(x) and integrate the other term, since we have already found the antiderivative (1/cos(x));

$u=\sin(x)\;\;\;\;\mathrm{d}v=\frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x$ $\mathrm{d}u=\cos(x)\,\mathrm{d}x\;\;\;\;v=\frac{1}{\cos(x)}$

Remember that

$\int u\,\mathrm{d}v=uv-\int v\,\mathrm{d}u$

Therefore,

$\frac{\sin(x)}{\cos(x)}-\int \cos(x)\cdot \frac{1}{\cos(x)}\,\mathrm{d}x$ $\frac{\sin(x)}{\cos(x)}-\int 1\,\mathrm{d}x$ $\frac{\sin(x)}{\cos(x)}-x$

This is the result of

$\int \frac{\sin^2(x)}{\cos^2(x)}\,\mathrm{d}x$

but we also have that x from the first part of the integration:

$\int \sec^2(x)\,\mathrm{d}x=x+\int \frac{\sin^2(x)}{\cos^2(x)}\,\mathrm{d}x=x+\frac{\sin(x)}{\cos(x)}-x=\frac{\sin(x)}{\cos(x)}$

sin(x)/cos(x) is tan(x); finally,

$\int \sec^2(x)\,\mathrm{d}x=\tan(x)+C$

Second method

$\int \sec^2(x)\,\mathrm{d}x$

This time we are going to let u be sec(x), so that sec²(x) becomes u²:

$u=\sec(x)\;\;\;\;\;\;\mathrm{d}u=\frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x$

sec(x) is 1/cos(x), so in order to differentiate it we apply the reciprocal rule:

$\frac{\mathrm{d} }{\mathrm{d} x }\frac{1}{f(x)}=-\frac{f'(x)}{[f(x)]^2}$

Therefore,

$\frac{\mathrm{d} }{\mathrm{d} x }\frac{1}{\cos(x)}=-\frac{-\sin(x)}{\cos^2(x)}=\frac{\sin(x)}{\cos^2(x)}$

Now, there is a problem: if we substitute we must integrate with respect to the new variable, u in our case, which means every function of the “old” variable x must be written in terms of u. We have

$\small u=\sec(x)=\frac{1}{\cos(x)}, {\color{Red} u^2}={\color{Red} \frac{1}{\cos^2(x)}} \,\,\text{and } \mathrm{d}u=\frac{\sin(x)}{\cos^2(x)}\,\mathrm{d}x=\sin(x)\cdot {\color{Red} \frac{1}{\cos^2(x)}}\,\mathrm{d}x$

The terms in red are simply u²:

$\small \mathrm{d}u=\sin(x)\cdot u^2\,\mathrm{d}x$

We’re almost there. How can we write sin(x) using cos²(x)? Remember the identity sin²(x)+cos²(x)=1:

$\small \sin^2(x)+\cos^2(x)=1\rightarrow \sin^2(x)=1-\cos^2(x)\rightarrow \sin(x)=\sqrt{1-\cos^2(x)}$

We know that u²=1/cos²(x), therefore cos²(x)=1/u²:

$\sin(x)=\sqrt{1-\cos^2(x)}\xrightarrow[]{\cos^2(x)=\frac{1}{u^2}}\sin(x)=\sqrt{1-\frac{1}{u^2}}$

This means

$\mathrm{d}u=u^2\sqrt{1-\frac{1}{u^2}}\,\mathrm{d}x$ $\mathrm{d}u=u^2\sqrt{\frac{u^2-1}{u^2}}\,\mathrm{d}x$ $\mathrm{d}u=\frac{u^2}{u}\sqrt{u^2-1}\,\mathrm{d}x$ $\mathrm{d}u=u\sqrt{u^2-1}\,\mathrm{d}x$

Let’s give a look at the integral:

$\int u^2\,\mathrm{d}x$

We want dx, therefore

$\mathrm{d}u=u\sqrt{u^2-1}\,\mathrm{d}x\rightarrow \mathrm{d}x=\frac{\mathrm{d}u}{u\sqrt{u^2-1}}$ $\int u^2\frac{\mathrm{d}u}{u\sqrt{u^2-1}}=\int \frac{u}{\sqrt{u^2-1}}\,\mathrm{d}u$

This is exactly what we wanted, no more x‘s. At this point what do you think we can do? Give it a try and then keep reading to see if you got it right.

This is what we are going to do: if you differentiate u²-1, you’ll get 2u, which will be the du but, since we want dx, it will “become” 1/2u and so it will cancel out with the u at the numerator. Since we used u before, we’ll change variable.

$t=u^2-1\;\;\;\;\;\;\mathrm{d}t=2u\,\mathrm{d}u\rightarrow \mathrm{d}u=\frac{\mathrm{d}t}{2u}$ $\int \frac{{\color{Red} u}}{\sqrt{t}}\frac{\mathrm{d}t}{2{\color{Red} u}}=\frac{1}{2}\int\frac{1}{\sqrt t}\,\mathrm{d}t=\frac{1}{2}\int t^{\frac{1}{2}}\,\mathrm{d}t$

1/2 is a constant and is moved in front of the integral; at this point we can apply the power rule for integrals:

$\frac{1}{2}\frac{t^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}=\frac{1}{2}\frac{t^{\frac{1}{2}}}{\frac{1}{2}}=t^\frac{1}{2}=\sqrt t$

We are almost done; let’s undo all substitutions. Remember:

$\left \{ u=\sec(x),\,t=u^2-1 \right \}\rightarrow t=\sec^2(x)-1; \sqrt t\xrightarrow[]{t=\sec^2(x)-1}\sqrt{\sec^2(x)-1}$

That’s the “final” result; “final” because we can actually do more: sec²(x)-1 comes from the trigonometric identity sec²(x)-tan²(x)=1 and, as you can see, it is equal to tan²(x); this means that

$\sqrt{\sec^2(x)-1}=\sqrt{\tan^2(x)}=\tan(x)$

Finally,

$\int \sec^2(x)\,\mathrm{d}x=\tan(x)+C$

That’it! We’ve integrated the function sec²(x) with two methods. If you weren’t able to do it by yourself, keep in mind that looking at the procedures is actually a great way to learn the mechanism and all of the techniques; this is how I started, so stay positive!

I really hope you found this post helpful and if you enjoyed it leave a like and subscribe for more! In the next post we’ll find the integral of csc²(x), so in the mean time, give it a try!

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Integral of sec²(x)|Two ways