Integrate secx/(1+cosecx)dx ?

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Feb 12, 2018

Answer:

#int secx/(1+cscx)*dx#

=#1/4Ln((1+sinx)/(1-sinx))+1/2*(1+sinx)^(-1)+C#

Explanation:

#int secx/(1+cscx)*dx#

=#int (1/cosx)/(1+1/sinx)*dx#

=#int sinx/[cosx*(1+sinx)]*dx#

=#int (sinx*cosx)/[(cosx)^2*(1+sinx)]*dx#

=#int (sinx*cosx*dx)/((1-(sinx)^2)*(1+sinx))#

=#int (sinx*cosx*dx)/((1-sinx)*(1+sinx)^2)#

After using #y=sinu# and #dy=cosu*du# transforms, this integral became

#int (y*dy)/((1-y)*(1+y)^2)#

Now, I decomposed integrand into basic fractions,

#y/((1+y)*(1-y)^2)=A/(1-y)+B/(1+y)+C/(1+y)^2#

After expanding denominator,

#A*(1+y)^2+B*(1-y^2)+C*(1-y)=y#

Set #x=-1#, #2C=-1#, so #C=-1/2#

Set #x=1#, #4A=1#, so #A=1/4#

Set #x=0#, #A+B+C=0#, so #B=1/4#

Hence,

#int (y*dy)/((1-y)*(1+y)^2)#

=#1/4int dy/(1-y)+1/4int dy/(1+y)-1/2int dy/(1+y)^2#

=#1/4Ln(1+y)-1/4Ln(1-y)+1/2*1/(1+y)+C#

=#1/4Ln((1+y)/(1-y))+1/2*(1+y)^(-1)+C#

Thus,

#int secx/(1+cscx)*dx#

=#1/4Ln((1+sinx)/(1-sinx))+1/2*(1+sinx)^(-1)+C#

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Dec 22, 2017

Answer:

#intsecx/(1+cscx)dx=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|+1/2 1/(1+sinx)+C#

Explanation:

Let

#I=intsecx/(1+cscx)dx#

Apply the substitution #x=2theta#:

#I=2int(sec2theta)/(1+csc2theta)d theta#

Apply the appropriate double-angle Trigonometric identities:

#I=2int(((1+tan^2theta)/(1-tan^2theta)))/(1+((1+tan^2theta)/(2tantheta)))d theta#

Rearrange:

#I=4int(tantheta/((1-tantheta)(1+tantheta)^3))sec^2thetad theta#

Apply the substitution #u=tantheta#:

#I=4intu/((1-u)(1+u)^3)du#

Apply partial fraction decomposition:

#I=1/2int(1/(1-u)+1/(1+u)+2/(1+u)^2-4/(1+u)^3)du#

Integrate directly:

#I=1/2{-ln|1-u|+ln|1+u|-2/(1+u)+2/(1+u)^2}+C#

Rearrange:

#I=1/2ln|(1+u)/(1-u)|-u/(1+u)^2+C#

Reverse the substitutions:

#I=1/2ln|(1+tan(x/2))/(1-tan(x/2))|-tan(x/2)/(1+tan(x/2))^2+C#

Apply the appropriate Trigonometric identities:

#I=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|-(sin(x/2)cos(x/2))/(cos(x/2)+sin(x/2))^2+C#

Rearrange:

#I=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|-1/2(2sin(x/2)cos(x/2))/(1+2sin(x/2)cos(x/2))+C#

Apply the appropriate double-angle Trigonometric identity:

#I=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|-1/2sinx/(1+sinx)+C#

Rearrange:

#I=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|+1/2 1/(1+sinx)-1/2+C#

Redefine #C# to remove the constant:

#I=1/2ln|(1+cosx+sinx)/(1+cosx-sinx)|+1/2 1/(1+sinx)+C#

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