# Integrate ${\sec 4x}$

How do I go about doing this? I try doing it by parts, but it seems to work out wrong:

\eqalign{ & \int {\sec 4xdx} \cr & u = \sec 4x \cr & {{du} \over {dx}} = 4\sec 4x\tan 4x \cr & {{dv} \over {dx}} = 1 \cr & v = x \cr & \int {\sec 4xdx} = x\sec 4x - \int {4x\sec 4x\tan 4xdx} \cr & \int {4x\sec 4x\tan 4xdx} : \cr & u = 4x \cr & {{du} \over {dx}} = 4 \cr & {{dv} \over {dx}} = \sec 4x\tan 4x \cr & v = {1 \over 4}\sec x \cr & \int {4x\sec 4x\tan 4xdx} = x\sec 4x - \int {\sec 4x} dx \cr & \int {\sec 4xdx} = x\sec 4x - \left( {x\sec 4x - \int {\sec 4xdx} } \right) \cr}

I don't know where to go from here, everything looks like it equals 0, where have I went wrong? Thank you!

EDIT: Is there an easier way to do this?

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Let $4x = t$, i.e., $4dx = dt$. $$I=\int \sec(4x) dx = \dfrac14\int \sec(t) dt = \dfrac14\int \dfrac{dt}{\cos(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{\cos^2(t)} = \dfrac14 \int \dfrac{\cos(t) dt}{1-\sin^2(t)}$$ Now let $\sin(t) = y$, to get $\cos(t) dt = dy$. Hence, we get that $$I = \dfrac14 \int \dfrac{dy}{1-y^2} = \dfrac18\left(\int\dfrac{dy}{1+y} + \int\dfrac{dy}{1-y}\right) = \dfrac18\left(\log(\vert 1+ y \vert) - \log(\vert 1 - y \vert)\right)+\text{const}$$ This gives us $$I = \dfrac18 \log \left(\left\vert \dfrac{1+y}{1-y}\right\vert \right) + \text{const} = \dfrac18 \log \left(\left \vert \dfrac{1+\sin(t)}{1-\sin(t)}\right \vert\right)+\text{const} = \color{red}{\dfrac18 \log \left(\left \vert \dfrac{1+\sin(4x)}{1-\sin(4x)}\right \vert\right)+\text{const}}$$

EDIT

You can simplify it further as much as you like.

\begin{align} \dfrac{1+\sin(4x)}{1-\sin(4x)} & = \dfrac{1+\sin(4x)}{1-\sin(4x)} \times \dfrac{1+\sin(4x)}{1+\sin(4x)} = \dfrac{(1+\sin(4x))^2}{1-\sin^2(4x)}\\ & = \dfrac{(1+\sin(4x))^2}{\cos^2(4x)} = (\sec(4x) + \tan(4x))^2 \end{align}

Hence, $$\boxed{\color{blue}{I = \dfrac{\log(\vert \sec(4x) + \tan(4x)\vert)}4 + \text{const}}}$$

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The answer is: $${1 \over 4}\ln |\sec 4x + \tan 4x| + C$$ maybe this is the same answer in a different form? – seeker May 17 '13 at 20:54
@Assad Updated to get it to the form you have. – user17762 May 17 '13 at 21:04

First try $u = 4x$. Then multiply numerator and denominator by $(\tan u + \sec u)$

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Nice hint ++++1 – Amzoti May 18 '13 at 0:48

The standard trick is $$\sec(4x)=\frac{(\sec 4x)(\sec 4x +\tan 4x )}{\sec 4x + \tan 4x}=\frac{\sec^2 4x+\sec 4x \tan 4x}{\sec 4x + \tan 4x}$$

Now take $u=\sec 4x + \tan 4x$, we have $du=4 \sec 4x \tan 4x + 4 \sec^2 4x$, so $\sec 4x dx=\frac{du}{4u}$.

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Use two substitutions. The first substitution transforms the integrand into $\sec \theta$, whose evaluation was asked here. The second substitution is the Weirstrass substitution. (See comment below). In the present case the second integral becomes an easy table integral:

$$\begin{eqnarray*} \int \sec 4x\,dx &=&\frac{1}{4}\int \sec \theta \,d\theta ,\qquad \theta =4x \\ &=&\frac{1}{4}\int \frac{1}{\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}} \,dt,\qquad t=\tan \frac{\theta }{2} \\ &=&\frac{1}{2}\int \frac{1}{1-t^{2}}dt=\frac{1}{2}\operatorname{arctanh}t+C \\ &=&\frac{1}{2}\operatorname{arctanh}\left( \tan \frac{\theta }{2}\right) +C \\ &=&\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) +C. \end{eqnarray*}$$

This integral can be rewritten as

$$\frac{1}{2}\operatorname{arctanh}\left( \tan 2x\right) =\frac{1}{4}\ln \left\vert \tan 2x+1\right\vert -\frac{1}{4}\ln \left\vert 1-\tan 2x\right\vert .$$

Comment: The Weierstrass substitution is a universal standard substitution to evaluate an integral of a rational fraction in $\sin \theta,\cos \theta$, i.e. a rational fraction of the form

$$R(\sin \theta,\cos \theta)=\frac{P(\sin \theta,\cos \theta)}{Q(\sin \theta,\cos \theta)},$$

where $P,Q$ are polynomials in $\sin \theta,\cos \theta$

$$\begin{equation*} \tan \frac{\theta }{2}=t,\qquad\theta =2\arctan t,\qquad d\theta =\frac{2}{1+t^{2}}dt \end{equation*},$$

which converts the integrand into a rational function in $t$. We know from trigonometry (see this answer) that

$$\cos \theta =\frac{1-\tan ^{2}\frac{\theta }{2}}{1+\tan ^{2}\frac{ \theta}{2}}=\frac{1-t^2}{1+t^2},\qquad \sin \theta =\frac{2\tan \frac{\theta }{2}}{1+\tan ^{2} \frac{\theta }{2}}=\frac{2t}{1+t^2}.$$

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@AmerucoTavares Thank you, I wasn't aware of this substitution till now – seeker May 18 '13 at 19:39
@Assad You are welcome. – Américo Tavares May 18 '13 at 19:43