How can I simplify this integral using u-substitution for BC Calculus?

[nyyfan0729]

I broke up the inegral of tan^(7)(theta)*sec^(5)(theta) into tan^(5)(theta)(sec^(2)(theta))(sec^(5)(theta). WHAT DO I DO NEXT

 

[arildno]

You have broken it up incorrectly.

 

[VietDao29]

I broke up the inegral of tan^(7)(theta)*sec^(5)(theta) into tan^(5)(theta)(sec^(2)(theta))(sec^(5)(theta). WHAT DO I DO NEXT

As arildno have pointed out, you've broken it incorrectly.
$$\tan ^ 5 \theta \sec ^ 2 \theta \sec ^ 5 \theta = \tan ^ 5 \theta \sec ^ 7 \theta \neq \tan ^ 7 \theta \sec ^ 5 \theta$$
In this integral, by converting tangent function to sine, and cosine function, we have:
$$\int \tan ^ 7 \theta \sec ^ 5 \theta d \theta = \int \frac{\sin ^ 7 \theta}{\cos ^ {12} \theta} d \theta$$
Now the power of the sine function is odd, it's common to use the substitution: $$u = \sin \theta$$, then use the well-known Pythagorean identity: cos²x + sin²x = 1 (or you can rearrange it a bit to give: sin²x = 1 - cos²x), to solve the problem.
If the power of the cosine function is odd, then it's common to use the substitution: $$u = \cos \theta$$.
Can you go from here? :)

 

[Hurkyl]

The textbooks I've seen usually present an algorithm for doing tan * sec integrals directly.

(That's as much pointed at nyyfan0729 as it is at VietDao29)

 

[arildno]

Asa follow-up on Hurkyl's suggestion, remember that:
$$\frac{d}{dx}\tan(x)=\sec^{2}(x)$$

 

[CrankFan]

Asa follow-up on Hurkyl's suggestion, remember that:
$$\frac{d}{dx}\tan(x)=\sec^{2}(x)$$

Hmmm, I guess I'm missing something because to me it seems like it's more important to remember that $$\frac{d}{dx}\sec(x)=\sec(x)tan(x)$$

Also if you do this with sine & cosine, as VietDao suggested, then I'm pretty sure that you need to make $$u = \cos \theta$$ the substitution instead of $$u = \sin \theta$$

 

[arildno]

Well, if you set $$\frac{dv}{dx}=\tan^{7}(x)\sec^{2}(x)$$
then the original integral is easily integrated as follows:
$$\int\tan^{7}(x)\sec^{5}(x)dx=\frac{1}{8}\tan^{8}(x)\sec^{3}(x)-\frac{3}{8}\int\tan^{9}(x)\sec^{3}(x)dx=\frac{1}{8}\tan^{8}(x)\sec^{3}(x)-\frac{3}{80}\tan^{10}(x)\sec(x)+\frac{1}{120}\tan^{12}(x)+C$$
or something like that.

Hmm..did a make a mistake somewhere?

Aargh, seems that I did..

 

[VietDao29]

The textbooks I've seen usually present an algorithm for doing tan * sec integrals directly.

(That's as much pointed at nyyfan0729 as it is at VietDao29)

Yes, thanks,

Also if you do this with sine & cosine, as VietDao suggested, then I'm pretty sure that you need to make the substitution $$u = \cos \theta$$ instead of $$u = \sin \theta$$

Whoops, sorry. My bad...
What the hell was I thinking about when writing this?

 

[GCT]

it seems that CrankFan has a nice suggestion, try setting $$u=sec \theta$$ and simplifying it from there.