Why doesn't this approach work for $\int \sec^4 x\,dx$?

Question

I been trying to integrate $\sec^4$ , without much luck. But I don't entirely understand why my result is invalid and would like some feedback if possible.

I'm attacking the issue in the following way

$$ \int (\sec^2{x})^2dx = \int (\tan^2+1)^2dx $$

then, I put $u=\tan^2+1$ which means $x = \arctan(\sqrt{u-1})$, which allows me to do the following backwards substitution

$$ \int (\sec^2{x})^2dx = \int (\tan^2+1)^2dx = \int u^2 \frac{1}{2u\sqrt{u-1}} dx = \frac{1}{2}\int u (u-1)^{\frac{-1}{2}} dx $$

Now using integration by parts I get

$$\begin{align*} \int u (u-1)^{\frac{-1}{2}} dx &= \frac{1}{2}(2u(u-1)^{\frac{1}{2}} - \int 2(u-1)^{\frac{1}{2}})\\\\ &= \frac{1}{2}(2u(u-1)^{\frac{1}{2}}-\frac{4}{3}(u-1)^{3/2}) \\\\ &= \frac{1}{2}(2(\tan^2{x}+1)(\tan^2{x})^{\frac{1}{2}}-\frac{4}{3}(\tan^2{x})^{3/2}) \end{align*}$$

This however, seems to be incorrect. How come?

Answer 1

You have to be careful with solving the equations; solving the equation

$$ 1 + \tan^2 x = u $$

for $x$ in terms of $u$ actually has many possible values: they are the values

$$ \pi n \pm \arctan \sqrt{u-1} $$

where $n$ ranges over all integers.

Ultimately, we can ignore the $\pi n$ part when rewriting the integral, because $\tan(z + n \pi) = \tan(z)$ and $d(z + n \pi) = dz$, and these are the only sorts places $\pi n $ would appear in the equation. However, we mustn't forget the sign.

Rather than donig two problems separately, it is convenient to define a new varaible $s$ to be $1$ or $-1$ as appropriate, and

$$ x = \pi n + s \arctan \sqrt{u-1} $$

and $s$ can be treated as a constant (depending on what you imagine for the fine details either $s$ really is a constant or $s$ is "locally constant", but either way, $ds = 0$). In the final simplifications, observe that

$$ \tan x = s \sqrt{u - 1} $$

which lets you convert back into trig functions while still keeping the signs right. Ultimately, all the $s$'s will cancel out (because $s^2 = 1$), so we don't have to worry about it in writing the final answer.

Incidentally, in my opinion it's easier not to actually solve for $x$: we can manage the substitution just from the knowledge that

$$ \tan x = s \sqrt{u-1} $$

from which we can derive

$$ \sec^2 x \, dx = s \frac{1}{2} (u-1)^{-1/2} du $$

and, of course, the original equation you set up lets us substitute $\sec^2 x = u$.

Answer 2

Your result is almost valid.

Taking a different approach, using Integration by Parts from the start:

Note that $\sec^2x = \frac{d}{dx} (\tan x)$.

We can use integration by parts:

$u = \sec^2 x \implies\,du = 2\sec^2 x \tan x\,dx$

$dv = \sec^2 x \,dx\implies \, v = \tan x$.

$$\int \sec^4 x \,dx = \sec^2x\tan x - 2\int \tan^2 x \sec^2 x$$

Now, the remaining integral can be easily solved by substitution: $w = \tan x,\;\implies dw = \sec^2 x$: $$2\int \tan^2 x \sec^2 x \,dx = 2\int w^2 \,dw = \dfrac 23 w + c$$

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Putting the above together gives us:

$$\int \sec^4 x \,dx = \sec^2x\tan x - \frac 23 \tan^3 x + C$$

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Note: Your answer is very, very close, and can be manipulated algebraically to closely match the above:

$$\frac{1}{2}\Big(2(\underbrace{\tan^2{x}+1}_{\sec^2 x})\underbrace{(\tan^2{x})^{\frac{1}{2}}}_{|\tan x|}-\frac{4}{3}\underbrace{(\tan^2{x})^{3/2}}_{|\tan^3 x|}\Big) = \sec^2 |\tan x| - \frac 23 |\tan^3 x| + C$$

Answer 3

I think your approach is not direct enough.

$\sec^2x \,dx = d(tan x)$

$\int \sec^4 x \,dx$

$= \int sec^2x\sec^2x \,dx$

$= \int sec^2x \,d(tan x)$

$= \int (1 + tan^2x)\,d(tan x)$

$= \int d(tan x) + \int tan^2x \,d(tan x)$

$= ...$