I think that my integral is wrong

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Robert Mak
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Integral of ((Tan x)^6)dx

i think that the solution is:

(1/5)(tan x)^5 + (1/3)(tan x)^3 - (2/3)(tan x)^3 + tan x - x + c

but if i derivate this solution it lead me to that:


(tan x)^4((tan x)^2 + 2)

?:confused:
 
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Yes, your integral is incorrect; however, try expanding the expression and then integrating.
 
I think your integral is correct. It must be the differentiation that is causing the problem. But no one can tell you why until you show your work.
 
It looks correct to me, also (although, why you haven't collected those terms in tan^3x, I don't know!)
 
I think he might forgetting to use the Chain Rule.
 
Thanks people; i did a mistake of signs lol.
 
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...
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