## jaersyn 3 years ago integrate (secx)^4 * sqrt (tanx) dx

1. Outkast3r09

I think you have to use Trig sub

2. Outkast3r09

or you might be able to do this

3. Outkast3r09

this is just a test idk if it will work $\int sec^2 x* sec^2 x*\sqrt{tanx} dx$

4. Outkast3r09

$cos^2 x + sin^2 x= 1$ $1+tan^2 x=sec^2 x$

5. jaersyn

you're correct, i'm close, but not exactly the ansewr

6. Outkast3r09

$\int sec^2 x * (tan^2 x + 1) * \sqrt{tan x}dx$

7. jaersyn

not trig sub, u-sub

8. jaersyn

yeah then let u = tanx; du = (secx)^2

9. Outkast3r09

you could use trig subs

10. Outkast3r09

trig sub will work with any trig integral... it's just a pain for something that can be done easier anyways $\int sec^2 x *(tan^{3/2} x+ \sqrt{tan^{1/2}x})$

11. Outkast3r09

$\int sec^2 x*tan^{3/2}x dx + \int sec^2 x*tan^{1/2}xdx$

12. Outkast3r09

$\int sec^2 x* (tanx)^{3/2} dx + \int sec^2 x*(tanx)^{1/2}$

13. Outkast3r09

let $u=tanx$ $du=sec^2 x$ $\int u^{3/2}du+\int u^{1/2}du$

14. jaersyn

i see where i went wrong

15. Outkast3r09

ok =]

16. jaersyn

ty

17. jaersyn

wait

18. jaersyn

∫sec2x∗(tan3/2x+tan1/2x−−−−−−√)

19. jaersyn

how does tan^2 (x) * tan^(1/2) x = tan^(3/2) (x)?

20. Outkast3r09

ehh it's supposed to be a 5/2

21. Outkast3r09