Method of integration? Integration by Parts

  • Thread starter Thread starter mathor345
  • Start date Start date
  • Tags Tags
    Integration Method
mathor345
Messages
16
Reaction score
0

Homework Statement



Evaluate. This may not require integration by parts:

integral of x^5 sec(x^6) dx

Homework Equations



integral sec x dx = ln | sec x + tan x| + C

integral u dv = uv - integral v du

... tabular integration process

The Attempt at a Solution



u = sec x^6
du = ln | sec x^6 + tan x^6 | + C
v = 1/6x^6
dv = x^5

secx^6 * 1/6x^6 - integral 1/6x^6 * ln |sec x^6 + tan x^6 |

(secx^6)/6 * x^6 - integral (x^6 ln | sec x^6 + tan x^6 | + C)/6

... this just looks really wrong. Help! :)
 
Physics news on Phys.org
Substitute u=x^6 first. It'll be a pleasant surprise:smile:
 
As a side note, the problem sort of <warned> you about not applying the part integration method. Yet you did and ended up nowwhere. That should tell you about what to do next.

As for this part:

u = sec x^6
du = ln | sec x^6 + tan x^6 | + C


it's wrong.
 
Last edited:
losiu99 said:
Substitute u=x^6 first. It'll be a pleasant surprise:smile:

Doh! I saw the answer as soon as I set that. Too much late night math homework after working full time really eats away at the brain, as we all see by the "backwards integration" that bigubau pointed out.

Thanks guys :)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top