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

1 Answer

cosxcos+sinxdx=12(x+ln(cosx+sinx))+c

Explanation:

Let

I=11+tanxdx

I=cosxcos+sinxdx

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

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

Equating the coefficients of cosx sinx and constants

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

l=12,n=12

cosx=12(cosx+sinx)+12(sinx+cosx)+0

I=cosxcos+sinxdx

I=12cosx+sinxcosx+sinxdx+12sinx+cosxcosx+sinxdx

I=12(I1+I2)

where,

I1=cosx+sinxcosx+sinxdx=1dx=x

I1=x

I2=sinx+cosxcosx+sinxdx

let

t=cosx+sinx

dtdx=sinx+cosx

dt=(sinx+cosx)dx

sinx+cosxcosx+sinxdx=(sinx+cosx)dxcosx+sinx=dtt=lnt

I2=ln(cosx+sinx)

I=12(I1+I2)

cosxcos+sinxdx=12(x+ln(cosx+sinx))+c