# Integral

#### Jameson

Staff member
I haven't worked out the solution yet, but given the answer and the general form of the integral I would guess trig substitution is the way to solve it. Question though - should the $(1-k^2)$ in the denominator actually be $(1-k^2)^2$?

If that's not the way to solve it then maybe there's something with the substitution $(k \cot(\theta))^2=k^2 \cot^2(\theta)=k^2(\csc^2(\theta)-1)$, but I'm not sure yet. Will post back if anything comes to mind.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hint :

$$\displaystyle \frac{d}{dx} \left(\cos^{-1} (f) \right) = \frac{-f'}{\sqrt{1-f^2}}$$​

#### MarkFL

Staff member
Hint: Try the substitution:

$$\displaystyle \cot(\theta)=\frac{\sqrt{1-k^2}}{k}\cos(u)$$

Hence:

$$\displaystyle \csc^2(\theta)\,d\theta=\frac{\sqrt{1-k^2}}{k}\sin(u)\,du$$

And the result will follow. #### ZaidAlyafey

##### Well-known member
MHB Math Helper
Hint: Try the substitution:

$$\displaystyle \cot(\theta)=\frac{\sqrt{1-k^2}}{k}\cos(u)$$

Hence:

$$\displaystyle \csc^2(\theta)\,d\theta=\frac{\sqrt{1-k^2}}{k}\sin(u)\,du$$

And the result will follow. No need for substitution. Of course it is more advisable on an elementary level .

#### FilipVz

##### New member
Thank you,

i solved it, using substitution:

$$\displaystyle u= (k*ctgθ)/(1-k^2 )$$

But, my result is:

$$\displaystyle ϕ=-arcsin((k∙ctgθ)/√(1-k^2 ))+c_2$$

$$\displaystyle ϕ=arccos((k∙ctgθ)/√(1-k^2 ))+c_2$$

Is this correct?

#### MarkFL

Staff member
Thank you,

i solved it, using substitution:

$$\displaystyle u= (k*ctgθ)/(1-k^2 )$$

But, my result is:

$$\displaystyle ϕ=-arcsin((k∙ctgθ)/√(1-k^2 ))+c_2$$

$$\displaystyle ϕ=arccos((k∙ctgθ)/√(1-k^2 ))+c_2$$

Is this correct?
Yes, that's another possible form. Consider the identity:

$$\displaystyle \sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}$$

along with the fact that the sum of an arbitrary constant and another constant is still an arbitrary constant. #### ZaidAlyafey

##### Well-known member
MHB Math Helper
$$\displaystyle \int \frac{-1}{\sqrt{1-x^2}} \, dx = \cos^{-1}(x)+A$$

$$\displaystyle -\int \frac{1}{\sqrt{1-x^2}} \, dx = -\sin^{-1}(x)+B$$

And this simply because

$$\displaystyle \cos^{-1}(x)+\sin^{-1}(x) =\frac{\pi}{2}$$