Question

How do you find the integral of `int 1/(1 + cot(x))`?

Answer 1

`-1/2(ln(abs(sin(x)+cos(x)))-x)+C`

Explanation:

Another method is to write this using all tangents:

`I=int1/(1+cot(x))dx=inttan(x)/(tan(x)(1+cot(x)))=inttan(x)/(tan(x)+1)dx`

Now, since all we have are tangents, we need a `sec^2(x)` in order to substitute. We can multiply the fraction by `sec^2(x)` in the numerator, but express it as `tan^2(x)+1` in the denominator (they're equal through the Pythagorean identity).

`I=int(tan(x)sec^2(x))/((tan(x)+1)(tan^2(x)+1))dx`

Letting `u=tan(x)` so that `du=sec^2(x)dx`, we see that:

`I=intu/((u+1)(u^2+1))du`

Now, we have to perform partial fraction decomposition:

`u/((u+1)(u^2+1))=A/(u+1)+(Bu+C)/(u^2+1)`

Multiplying through:

`u=A(u^2+1)+(Bu+C)(u+1)`

`u=Au^2+A+Bu^2+Bu+Cu+C`

Factor in three groups: those with `u^2`, those with `u`, and constants.

`u=u^2(A+B)+u(B+C)+(A+C)`

`color(purple)0u^2+color(red)1u+color(brown)0=u^2color(purple)((A+B))+ucolor(red)((B+C))+color(brown)((A+C))`

Comparing the two sides, we see that:

`{(A+B=0),(B+C=1),(A+C=0):}`

Subtracting the second equation from the third, we see that `A-B=-1`. Adding this to the first equation shows that `2A=-1` and `A=-1/2`. It then follows that:

`{(A=-1/2),(B=1/2),(C=1/2):}`

So:

`u/((u+1)(u^2+1))=1/2(1/(u+1))+1/2((u+1)/(u^2+1))`

Returning to the integral now:

`I=-1/2int1/(u+1)du+1/2int(u+1)/(u^2+1)du`

`I=-1/2int1/(u+1)du+1/2intu/(u^2+1)du+1/2int1/(u^2+1)du`

Modifying the second integral slightly:

`I=-1/2int1/(u+1)du+1/4int(2u)/(u^2+1)du+1/2int1/(u^2+1)du`

Now all three integrals can be integrated rather painlessly:

`I=-1/2ln(abs(u+1))+1/4ln(abs(u^2+1))+1/2arctan(u)`

`I=-1/2ln(abs(tan(x)+1))+1/4ln(tan^2(x)+1)+1/2arctan(tan(x))`

`color(blue)(I=-1/2ln(abs(tan(x)+1))+1/4ln(sec^2(x))+1/2x`

This is a fine final answer, once the constant of integration is added, but we can fiddle around a little more to achieve some fun simplification.

`I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))+1/2(1/2ln(sec^2(x)))+1/2x`

Rather sneakily, bring one of the `1/2`s outside the `ln(sec^2(x))` in, effectively using the `log(a^b)=blog(a)` rule in reverse. (Absolute value bars will be added since we've just taken the square root:)

`I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))+1/2ln(abs(sec(x)))+1/2x`

Now we can bring a `-1` in as a `-1` power to make `sec(x)` into `cos(x)`:

`I=-1/2ln(abs((sin(x)+cos(x))/cos(x)))-1/2ln(abs(cos(x)))+1/2x`

Factor `-1/2` and use the rule that `log(A)+log(B)=log(AB)`:

`I=-1/2(ln(abs((sin(x)+cos(x))/cos(x)))+ln(abs(cos(x)))-x)`

`color(green)(I=-1/2(ln(abs(sin(x)+cos(x)))-x)+C`

Answer 2

`1/2x-1/2ln|sinx+cosx|+C`.

Explanation:

Let `I=int1/(1+cotx)dx=intsinx/(sinx+cosx)dx`.

Recall that, `d/dx(sinx+cosx)=cosx-sinx,` so that,

`I=1/2int(2sinx)/(sinx+cosx)dx`

`=1/2int{(sinx+cosx)-(cosx-sinx)}/(sinx+cosx)dx`

`=1/2int(sinx+cosx)/(sinx+cosx)dx-1/2int(cosx-sinx)/(sinx+cosx)dx`

`=1/2int1dx-1/2int{d/dx(sinx+cosx)}/(sinx+cosx)dx`

`:. I=1/2x-1/2ln|sinx+cosx|+C`.

Note that the later integral has been derived as a special case of

`"the Result : "int(f'(x))/f(x)dx=ln|f(x)+c.`

This useful Result can easily be proved by substituting `f(x)=t.`

Enjoy Maths.!