## AmTran_Bus one year ago Help me integrate tan^3x

1. AmTran_Bus

|dw:1434148868513:dw|

2. AmTran_Bus

|dw:1434148890008:dw|

3. AmTran_Bus

So now u sub?

4. anonymous

sec^2 x - 1 = tan^2 x

5. AmTran_Bus

Didn't I do that in my step above?

6. anonymous

Oh! sorry u did

7. anonymous

So now distribute. Do you know the antiderivative of $$\tan x$$?

8. AmTran_Bus

Yes. ln sec x +c

9. anonymous

Okay, so from distributing you should get $\int \tan x\sec^2x\,dx-\int\tan x\,dx=\int\tan x\sec^2x\,dx-\ln|\sec x|+C$ You mentioned a substitution, that's a good idea. What do you think might work?

10. AmTran_Bus

I would say u= tanx

11. anonymous

Yep, that should work. Another one you can use: write $$\tan x\sec x\sec x$$, then $$u=\sec x$$ will work too.

12. anonymous

$\int\limits \tan x ~dx=\int\limits \frac{ \sin x }{ \cos x }dx=-\ln \left| \cos x \right|$

13. AmTran_Bus

|dw:1434149575520:dw|

14. AmTran_Bus

Well, but I need to consider my du

15. anonymous

Right, if $$u=\tan x$$, then $$du=\sec^2x\,dx$$, yes?

16. AmTran_Bus

So it should say integral ( u du - rest) right?

17. anonymous

Yeah that's $\int u\,du-\ln|\sec x|+C$ and so on.

18. AmTran_Bus

Nice. Let me evaluate that and plug in u and see if it is right.

19. AmTran_Bus

Well, guess im stuck

20. AmTran_Bus

I know part of the answer has to be|dw:1434149823843:dw|

21. anonymous

If I had to guess what the problem was... Hint: Identities.

22. AmTran_Bus

Hum. I gave up and looked at the book. The answer is |dw:1434149984333:dw|

23. AmTran_Bus

But I just dont see it.

24. anonymous

$\int u\,du=\frac{1}{2}u^2+C$ by the power rule. You set $$u=\tan x$$, so this becomes $$\dfrac{\tan ^2x}{2}+C$$.

25. anonymous

I was thinking: $\begin{split} \int\tan^3(x)~dx &= \int \frac{\sin^3(x)}{\cos^3(x)}dx \\ &= \int \frac{1-\cos^2(x)}{\cos^3(x)}~d(-\cos(x)) &,&u=\cos(x) \\ &= -\int\left(\frac{1}{u^3}-\frac{1}{u}\right)~du \end{split}$

26. anonymous

Somehow it seems wrong though.

27. AmTran_Bus

@SithsAndGiggles ok.

28. AmTran_Bus

I forgot a basic step, you were right.

29. anonymous

@wio seems right to me. $-\int\frac{du}{u^3}+\int\frac{du}{u}=\frac{1}{2u^2}+\ln|u|+C=\frac{\sec^2x}{2}+\ln|\sec x|+C$