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I'm once again stuck; I'm trying to find $$\int{(\sec^2{x} \tan^2{x})dx}$$ but end up with things like: $\tan^2x-\int{2\tan^2x \sec^2x}$, which doesn't help.

Would it be best to approach using integration by parts or substitution?

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    What is the derivative of tangent? – 2017-01-06
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    Your approach would work. Can you please include it. Note that there are more efficient ways to do it. – 2017-01-06
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    @DavidP Why do you ask?.. – 2017-01-06

3 Answers 3

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Let $u=\tan x$, $du=\sec^2 x~dx$. The integral becomes:

$$\int u^2\ du$$

Can you continue from here?

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    How'd it become so easy? – 2017-01-06
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    Can you integrate $\int x^2 dx$? – 2017-01-06
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    @Tobi It isn't necessary for all integrals to be solved by parts. Some are straightforward and others via some poking about can be solved by parts. Analysis of the integral is very important before using a particular technique. – 2017-01-06
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    ZacharySelk has used substitution to simplify the integral. – 2017-01-06
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    My mathematical analysis is deplorable, I always take the most complicated route... – 2017-01-06
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    Integral calculus takes longer to develop a good feel for what technique is best in a particular situation. The more question you do on it the better your analysing will get. Its worth noting there is nothing wrong with doing it the long way other than it takes more time. You might also discover something new doing it less efficiently. – 2017-01-06
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I'm assuming you got your answer using 'by parts'. Below is how you could finish it using 'by parts'. Note that other solutions such as by @ZacharySelk are simpler.

Using your line of working:

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

You can move the $- 2\int \sec^2x \tan^2x dx$ to the left hand side of the equation by addition.

$$\int \sec^2x \tan^2x dx+ 2\int \sec^2x \tan^2x dx= tan^2x +c, c\in\mathbb{R}$$

Note that once we have a side without an integral on it you need to include a constant of integration. I have used $c$.

The two expressions on the left hand side are the same so you can add them giving:

$$3\int \sec^2x \tan^2x dx= tan^2x +c$$

So simply divide by 3 to get your answer:

$$\int \sec^2x \tan^2x dx= \frac{tan^2x}{3} +\frac{c}{3}$$

Note that as $c$ is an real number we could replace $\frac{c}{3}$ with $c_2$ to write the answer more neatly as:

$$\int \sec^2x \tan^2x dx= \frac{tan^2x}{3} +c_2$$

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    Genius........! – 2017-01-06
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    This approach is used when integrating something like $\int e^x\sin x dx$. Use 'by parts' twice then move the integral from one side to the other and divide by a number. – 2017-01-06
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    To be honest I should have spotted that it followed the rule: $\int{a' \cdot a^n} = k(a^{n+1})$. The question would've been solved in maybe a minute... – 2017-01-07
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Putting $u = \tan x$, $du=\sec^2x\ dx$, we get:

$\ \ \ \ \int u^2du$

$=\frac{u^3}{3}+c$

$=\frac{\tan^3x}{3}+c$