## farukk 2 years ago integrate tan^3 xsec x dx...!

1. farukk

$\int\limits \tan^3 x secx dx$

2. farukk

what to do?

3. experimentX

$\tan^3 x \sec x = \tan x \sec x ( \sec^2 x - 1)$ seems like it will get the same

4. farukk

yeah ..next step

note that d(sec x) =sec x tan x dx

6. farukk

yes all right

7. farukk

what's next..

let u = sec x then proceed. du= sec x tan x dx so it all becomes $$\int (u^2 - 1) du$$

9. farukk

dear are you solving integral by parts?

10. farukk

or by substituion?

11. farukk

it will become like this i guess u^3/3 - u +c

$$\int (\sec^2-1) \tan x \sec x dx$$ =$$\int (\sec^2-1) d(\sec x)$$ =$$\int (u^2-1) du$$, where u=\sec x\) by substitution.

13. farukk

yes next step..

eh? $$\int u^n du= \frac{u^{n+1}}{n+1} +C$$

15. farukk

after applying we ll get as i above wrote ?

you'll get the answer....from integrating the above eq.

17. farukk

wait let me write what i am getting.

18. farukk

$\frac{ \sec^3x }{ 3 }- secx +c = answer ???$

it should be, yeah. :) you can always check your answers with wolfram.

20. farukk

wolfram.?

22. farukk

but answer is differnet in my book?

23. farukk

$\frac{ 1 }{ 3 }(secxtan^2x-2secx)+c =answer ( book)$

24. farukk

just take sec x out. and also, the integral, $$\int du=u$$

26. farukk

did't get..

$$\int (u^2-1) du = \int u^2 du - \int du$$ following the integral, it becomes, $$\frac{u^3}{3} - u+C$$

28. farukk

i have already written this.Scroll up a bit

29. farukk

i just wanna get the same answer as in my book...:)

oh. unfortunately, either your book's wrong, or the computer integration is wrong. because wolfram agrees with you....lol

31. farukk

yeah i have seen there but what you think about book's anwer is that wrong,there is really less possibilit though and yes lolz?

32. farukk

i have three books with same answer(:

lol i dun think wolfram's wrong, in anycase....lol

34. farukk

if we solve this by ''integration by parts'' rather tha by substituion?

35. farukk

$\int\limits uvdx= u \int\limits vdx - \int\limits(\int\limits vdx) u'dx$

same way, but this time i still let du=d sec x tan x dx. and v=sec^2 -1 sorry, gtg, but i think you'll still get the same answer...(or that's what my calc tells me)

37. farukk

@Shadowys okay i think my all books are wrong this time.(there is no benifit or edge having more than one books) :)!

38. farukk

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