Question

Integrate 1/(a+bcotx)?

Answer 1

See below.

Explanation:

1) I converted this integral in terms of `tanx`

2) I used `u=tanx` and `du=[(tanx)^2+1]*dx` for converting integrand into fraction in terms of `u`

3) I used basic fractions method.

4) I rewrote `u=tanx` for finding solution.

`int dx/(a+b*cotx)`

`int (tanx*dx)/(a*tanx+b)`

`int (tanx)*([(tanx)^2+1]*dx)/[(a*tanx+b)*[(tanx)^2+1]]`

After using `u=tanx` and `du=[(tanx)^2+1]*dx` transforms, this integral became,

`int (u*du)/[(u^2+1)*(au+b)]`

=`int ((bu+a)du)/[(u^2+1)*(a^2+b^2)]`-`int (ab*du)/[(au+b)*(a^2+b^2)]`

`b/(2a^2+2b^2)*ln(u^2+1)+a/(a^2+b^2)*arctanu-b/(a^2+b^2)*ln(au+b)+C`

`b/(2a^2+2b^2)*ln[(tanx)^2+1]+a/(a^2+b^2)*arctan(tanx)-b/(a^2+b^2)*ln(atanx+b)+C`

`b/(2a^2+2b^2)*ln[(secx)^2]+a/(a^2+b^2)*x-b/(a^2+b^2)*ln(atanx+b)+C`

`b/(a^2+b^2)*ln(secx)+a/(a^2+b^2)*x-b/(a^2+b^2)*ln(atanx+b) +C`

Answer 2

`1/(a^2+b^2){ax-bln|(asinx+bcosx)|}+C.`

Explanation:

Denote, `int1/(a+bcotx)dx," by, "I." Then, "I=intsinx/(asinx+bcosx)dx.`

Evibently, `a^2+b^2ne0.`

A useful technic to solve this type of integrals, is to decompose the

Nr. as, `"Nr.="l*Dr.+m*d/dx(Dr.), where, l,m in RR....(star).`

So, let, for some `l,m in RR,`

`sinx=l(asinx+bcosx)+md/dx(asinx+bcosx)`

`:.sinx=l(asinx+bcosx)+m(acosx-bsinx), i.e.,`

`1sinx+0cosx=(la-mb)sinx+(lb+ma)cosx.`

`:. la-mb=1, &, lb+ma=0.` Solving these for `l,m;` we get,

`l=a/(a^2+b^2), m=-b/(a^2+b^2)...........(star')`

Now, `(star) rArr sinx/(asinx+bcosx)=l+{md/dx(asinx+bcosx)}/(asinx+bcosx).`

`:. I=int[l+{md/dx(asinx+bcosx)}/(asinx+bcosx)]dx,`

`=l*int1dx+m*int{(d/dx(asinx+bcosx))/(asinx+bcosx)}dx,`

`=lx+mln|(asinx+bcosx)|.`

Finally, using `(star'),` we have,

`I=a/(a^2+b^2)*x-b/(a^2+b^2)*ln|(asinx+bcosx)|, or,`

`I=1/(a^2+b^2){ax-bln|(asinx+bcosx)|}+C.`

Foot Note :

Observe that, the above Solution is based on the valid

condition that `a^2+b^2 ne0.`

I hope that the Questioner will work out the Solution for

`a=0 ( &, :., bne0), and, b=0; (ane0.)`

Enjoy Maths.!