Solve Differential Equations with Ease - No More Struggles!

deteam
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Differential Equations PLEASE!

**problem solved**
 
Last edited:
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It seems that you didn't multiply tan(x) with the -cos(x) term.
 


nicksauce said:
It seems that you didn't multiply tan(x) with the -cos(x) term.

i turned tan(x) into sinx/cosx and canceled out the cosx, and was left with

-sin^2 x + cos^2 x + sinx + sin^2 x -cosx = cos^2 x

any ideas?
 


Like I said, you wrote:

-sin^2 x + cos^2 x +sinx +[tan x] sinxcosx -cosx = cos^2x

When it should be:

-sin^2 x + cos^2 x +sinx +[tan x]( sinxcosx -cosx) = cos^2x
 


nicksauce said:
Like I said, you wrote:

-sin^2 x + cos^2 x +sinx +[tan x] sinxcosx -cosx = cos^2x

When it should be:

-sin^2 x + cos^2 x +sinx +[tan x]( sinxcosx -cosx) = cos^2x

aww, such a simple mistake, thanks a lot :)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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