## suju101 3 years ago integrate tan^3(x/4)*sec^2(x/4) from-2/3pi to pi help with the steps plz.

1. amistre64

since sec^2 is the derivaive of tan, this is a simple enough arrangement: $\int f(u)du$

2. amistre64

i thinkk you might be off by a 4 tho

3. amistre64

x/4 pops out a spurious little 1/4 that missing

4. shinigami1m

is -2pi/3 to pi = to pi to -2pi/3 ???

5. amistre64

then given the interval [a,b] you work it out: F(b)-F(a)

6. shinigami1m

@_@

7. shinigami1m

|dw:1326817635699:dw|

8. amistre64

$[tan^4(\frac{1}{4}x)]'=\frac{1}{4}tan^3(\frac{1}{4}x)sec^2(\frac{1}{4}x)$ so we have to multiply thru by 4 to begin with

9. amistre64

|dw:1326817802065:dw|

10. TuringTest

$u=\frac{x}{4}\to du=\frac{dx}{4}\to4du=dx$$\int_0}{|dw:1326817852162:dw| 11. suju101 \[\Pi ^{2}\div9{(65\Pi ^{2}/9)-26)}$ i got this ans. s this correct??i

12. amistre64

lol, I just derived tan^4(x/4) and got a spurious 1/4 so that cant be right

13. amistre64

hmmm, im off by something

14. amistre64

wolf agrees with turing

15. amistre64

oh yeah, the 4 pops out to get rid of the 1/4 lol

16. shinigami1m

derivative of (x/4) i think u miss it during derivative

17. TuringTest

yep, one comes out on top and bottom, so they cancel

18. amistre64

a was using an abacus ;)

19. TuringTest

lol

20. suju101

|dw:1326818150512:dw| i did this process n got that ans

21. TuringTest

is the problem$\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx$?

22. suju101

yes

23. TuringTest

$\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx$$u=\frac x4\to du=\frac{dx}{4}\to4du=dx$so...$\int_{\frac{-2\pi}{3}}^{\pi}\tan^3(\frac x4)\sec^2(\frac x4)dx=4\int_{\frac{-\pi}{6}}^{\frac{\pi}{4}}\tan^3u\sec^2udu=4(\frac14\tan^4(\frac x4))|_{\frac{-2\pi}{3}}^{\pi}$$=\tan^4(\frac x4)|_{\frac{-2\pi}{3}}^{\pi}$evaluate...

24. robtobey

$\text{Tan}\left[\frac{\pi }{4}\right]^4-\text{Tan}\left[\frac{\frac{-2}{3}\pi }{4}\right]^4=\frac{8}{9}$

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