Homework Help: More trig substitution help

1. Jan 31, 2008

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

2. Jan 31, 2008

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
$\int \tan^{2}x \left(\sec^{2}x - 1\right) \mathrm{d}x = \int \tan^{2}x \sec^{2}x \mathrm{d}x - \int \tan^{2}x\mathrm{d}x$.

3. Feb 1, 2008

Gib Z

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

4. Feb 1, 2008

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
$$\int \sec^4 dx$$
?

5. Feb 1, 2008

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.

6. Feb 1, 2008

fermio

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

Last edited: Feb 1, 2008
7. Feb 1, 2008

Gib Z

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

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