Integrate 1/(a+bcotx)?

2 Answers

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#

Sep 26, 2017

# 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.!