Integration (1 Viewer)

[independantz]

I(secx)^3(tanx).dx
I(cosecx).dx
I(secx)^2.dx

Could someone please help me with these three questions, thanks

 

[Mark576]

int. sec³x.tanx.dx
Let u=secx
du/dx=secxtanx
dx=du/secxtanx
int. u².du
=u³/3 + c
=sec³x/3 + c

2. int. cosecx.dx
Multiply top and bottom by cosecx+cotx, then use u=cosecx+cotx.

3. int. sec²x = tanx + c

 

[vds700]

for 2, isn't it better to use the substitution t = tan(x/2)

I cosecx dx
= I (1/sinx) dx

let t = tan(x/2)
sinx = 2t/(1+t^2)
dx = 2dt/(1+ t^2)

=I(1+t^2)/2t . 2dt/(1+t^2)
=Idt/t
=lnt + c
=ln(tan(x/2)) + c

 

[Mark576]

Either method is appropriate I guess, using the substitution u=cosecx+cotx works out easier than you would think.

 

[Trebla]

Either way is fine, because it turns out that: 1/(cosec x + cot x) = tan (x/2)
* If you multiply it by (cosec x + cot x)/(cosec x + cot x), the answer is:
- ln (cosec x + cot x) + c = ln (1/(cosec x + cot x)) + c

* If you do it by t-formulae the answer is: ln (tan x/2) + c, which is equivalent.

I prefer the earlier one because it looks more clever lol...