Trig Substitution: Solving for the Missing Identity

[quickclick330]

[SOLVED] More trig substitution help...

I've looked at this problem about 3 times and still can't figure it out...where identity did they use to substitute out the part in the red box? Thanks for the help

 

[penguino]

What's the step you don't understand?
$$\int \tan^{4}x \mathrm{d}x = \int \tan^{2}x \left(\sec^{2}x - 1\right) \mathrm{d}x$$
or
$\int \tan^{2}x \left(\sec^{2}x - 1\right) \mathrm{d}x = \int \tan^{2}x \sec^{2}x \mathrm{d}x - \int \tan^{2}x\mathrm{d}x$.

 

[Gib Z]

$$\sec^2 x - 1 = \tan^2 x$$ and $$\tan^2 x \cdot \tan^2 x = \tan^4 x$$.

 

[fermio]

for me it's unclear how to integrate
$$\int\tan^2 x\sec^2 x dx=\int (\sec^2 x-1)\sec^2 x dx=\int \frac{1}{\cos^4}dx-\tan x$$
So how to integrate
$$\int \sec^4 dx$$
?

 

[Gib Z]

Integration by parts a few times does it, or write

$$\int \sec^2 x (\tan^2 x + 1) dx$$ and let u= tan x.

But rather than integrate sec^4, keep the original integral,
$$\int \tan^2 x \sec^2 x dx = \int u^2 du$$ when u= tan x.

 

[fermio]

I see
$$\int\tan^2 x\sec^2 x dx=\int \tan^2 x d(\tan x)=\frac{1}{3}\tan^3 x+C$$

 

[Gib Z]

You forgot the x on the end of the tan, but other than that, its correct.