## spyros 3 years ago integral(sec(x)^3)

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1. experimentX

$\int \sec^3x dx \text{ or } \int \sec (x^3) dx ??$

2. spyros

$\int\limits(\sec(x)^3dx$

3. experimentX

former or latter??

4. spyros

former

5. experimentX

$\sec^3x = \sec^2 \sec x = (1 + \tan^2x) \sec x$

6. spyros

actually( $\int\limits{\sqrt(1+(x-1)^2)dx}$

7. spyros

@experimentX i'll try that one

8. spyros

i came to integral(sec(x)^3dx) from integral(sqrt(1+(x-1)^2)dx

9. slaaibak

$\int\limits_{}^{} \sec^3(x) = \int\limits_{}^{}\sec x {d \over dx} \tan x = \sec x \tan x - \int\limits \sec (x)\tan^2x$ $\int\limits \sec^3(x) = \sec x \tan x - \int\limits \sec x (\sec^2(x) - 1)$

10. spyros

ok, i get it

11. slaaibak

Now take the second part further

12. experimentX

int sqrt(1 + (x-1)^2) dx = (1/4*(2*x-2))*sqrt(2+x^2-2*x)+(1/2)*arcsinh(x-1)

13. experimentX

if case of any confusion plugin integration to wolframalpha and click for show steps

14. spyros

pellet

15. spyros

sh.it

16. Callisto

Just a little try, not sure if it is correct but hope it helps $\int sec^3xdx$$=\int secx(1+tan^2x)dx$$=\int secx+secxtan^2xdx$$=\int secxdx+\int secxtan^2xdx$$=\int \frac{secx(secx+tanx)}{secx+tanx}dx+\int tanxd(secx)$$=\int \frac{1}{secx+tanx}d(secx+tanx)+[secxtanx-\int secxd(tanx)]$$=\ln |secx+tanx|+[secxtanx-\int secxsec^2xdx] +C$$=\ln |secx+tanx|+[secxtanx-\int sec^3xdx] +C$ So, we've got $\int sec^3xdx=\ln |secx+tanx|+secxtanx-\int sec^3xdx +C$$2\int sec^3xdx=\ln |secx+tanx|+secxtanx +C$$\int sec^3xdx=\frac{1}{2}(\ln |secx+tanx|+secxtanx) +C$