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

Question

Christopher Allers · Accepted Answer

#intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c#

Let

#I=int1/(1+tanx)dx#

#I=intcosx/(cos+sinx)dx#

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

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

Combining constants and the coefficients of cosx and sinx

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

#l=1/2, n=1/2#

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

#I=intcosx/(cos+sinx)dx#

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

#I=1/2(I_1+I_2)#

where,

#I_1=int(cosx+sinx)/(cosx+sinx)dx=int1dx=x#

#I_1=x#

#I_2=int(-sinx+cosx)/(cosx+sinx)dx#

let

#t=cosx+sinx#

#dt/dx=-sinx+cosx#

#dt=(-sinx+cosx)dx#

#int(-sinx+cosx)/(cosx+sinx)dx=int((-sinx+cosx)dx)/(cosx+sinx)=intdt/t=lnt#

#I_2=ln(cosx+sinx)#

#I=1/2(I_1+I_2)#

#intcosx/(cos+sinx)dx=1/2(x+ln(cosx+sinx))+c#

Elijah Abram · Answer

undefined To find the integral of $ \frac{1}{1 + \tan(x)} $, you can use the substitution method. Let $ u = \tan(x) $. Then, $ du = \sec^2(x) \, dx $. Rewrite the integral in terms of $ u $ and $ du $, and integrate with respect to $ u $. After integrating, substitute back $ \tan(x) $ for $ u $ to obtain the final result. The integral of $ \frac{1}{1 + \tan(x)} $ is $ \ln|1 + \tan(x)| + C $, where $ C $ is the constant of integration.