How do you find the integral of int 1/(1 + tan(x))11+tan(x)?

1 Answer

intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+ccosxcos+sinxdx=12(x+ln(cosx+sinx))+c

Explanation:

Let

I=int1/(1+tanx)dxI=11+tanxdx

I=intcosx/(cos+sinx)dxI=cosxcos+sinxdx

cosx=l(cosx+sinx)+m(-sinx+cosx)+ncosx=l(cosx+sinx)+m(sinx+cosx)+n

cosx=(l+m)cosx+(l-m)sinx+ncosx=(l+m)cosx+(lm)sinx+n

Equating the coefficients of cosx sinx and constants

l+m=1, l-m=0, n=0l+m=1,lm=0,n=0

l=1/2, n=1/2l=12,n=12

cosx=1/2(cosx+sinx)+1/2(-sinx+cosx)+0cosx=12(cosx+sinx)+12(sinx+cosx)+0

I=intcosx/(cos+sinx)dxI=cosxcos+sinxdx

I=1/2int(cosx+sinx)/(cosx+sinx)dx+1/2int(-sinx+cosx)/(cosx+sinx)dxI=12cosx+sinxcosx+sinxdx+12sinx+cosxcosx+sinxdx

I=1/2(I_1+I_2)I=12(I1+I2)

where,

I_1=int(cosx+sinx)/(cosx+sinx)dx=int1dx=xI1=cosx+sinxcosx+sinxdx=1dx=x

I_1=xI1=x

I_2=int(-sinx+cosx)/(cosx+sinx)dxI2=sinx+cosxcosx+sinxdx

let

t=cosx+sinxt=cosx+sinx

dt/dx=-sinx+cosxdtdx=sinx+cosx

dt=(-sinx+cosx)dxdt=(sinx+cosx)dx

int(-sinx+cosx)/(cosx+sinx)dx=int((-sinx+cosx)dx)/(cosx+sinx)=intdt/t=lntsinx+cosxcosx+sinxdx=(sinx+cosx)dxcosx+sinx=dtt=lnt

I_2=ln(cosx+sinx)I2=ln(cosx+sinx)

I=1/2(I_1+I_2)I=12(I1+I2)

intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+ccosxcos+sinxdx=12(x+ln(cosx+sinx))+c