How do you integrate #1/(1+tanx) dx#?

Question

Scarlett Allison · Accepted Answer

#int1/(1+tanx)"d"x=1/4(2x+ln(1+sin2x))+"c"#

We want to find #int1/(1+tanx)"d"x#.

We start by substituting #u=tanx# and #"d"u=sec^2x"d"x=(1+tan^2x)"d"x=(1+u^2)"d"x#

So

#int1/(1+tanx)"d"x=int1/((1+u)(1+u^2))"d"u#

Now we perform a partial fraction decomposition on the integrand

#1/((1+u)(1+u^2))=A/(1+u)+(Bu+C)/(1+u^2)#

#1=A(1+u^2)+(Bu+C)(1+u)#

Letting #u=-1# gives

#1=A(1+(-1)^2)=2ArArrA=1/2#

Letting #u=0# gives

#1=1/2(1+0)+CrArrC=1/2#

Letting #u=1# gives

#1=1/2(2)+(B+1/2)(2)=2B+2rArrB=-1/2#

So,

#1/((1+u)(1+u^2))=1/2(1/(1+u)+(1-u)/(1+u^2))#

and

#int1/((1+u)(1+u^2))"d"x=#

#1/2(int1/(1+u)"d"u+int1/(1+u^2)"d"u-intu/(1+u^2)"d"u)=#

#1/2(lnabs(1+u)+arctanu-1/2ln(1+u^2))+"c"#

Reversing for #u=tanx# gives

#1/4(2lnabs(1+tanx)+2arctan(tanx)+ln(1/(1+tan^2x)))+"c"#

#=1/4(2x+ln(cos^2x+2sinxcosx+sin^2x))+"c"#

#=1/4(2x+ln(1+sin2x))+"c"#

Aurora Alvarado · Answer

# 1/2{x+ln|(cosx+sinx)|}+C#.

Let, #I=int1/(1+tanx)dx=intcosx/(cosx+sinx)dx#. #:. I=1/2int(2cosx)/(cosx+sinx)dx#, #=1/2int{(cosx+sinx)+(cosx-sinx)}/(cosx+sinx)dx#, #=1/2int{(cosx+sinx)/(cosx+sinx)+(cosx-sinx)/(cosx+sinx)}dx#, #=1/2int{1+(d/dx(cosx+sinx))/(cosx+sinx)dx}#, # rArr I=1/2{x+ln|(cosx+sinx)|}+C#.

Paisley Johnson · Answer

Use the substitution #tanx=u#.

Let #I=int1/(1+tanx)dx# Apply the substitution #tanx=u#: #I=int1/((1+u^2)(1+u))du# Apply partial fraction decomposition: #I=1/2int((1-u)/(1+u^2)+1/(1+u))du# Rearrange: #I=1/2int(1/(1+u^2)-1/2(2u)/(1+u^2)+1/(1+u))du# Integrate term by term: #I=1/2(tan^(-1)u-1/2ln(1+u^2)+ln(1+u))+C# Reverse the substitution: #I=1/2(x-ln(secx)+ln(1+tanx))+C# Simplify: #I=1/2(x+ln(sinx+cosx))+C#

Aurora Alvarado · Answer

undefined To integrate $ \frac{1}{1 + \tan(x)} \, dx $, you can use the substitution method.

1. Let $ u = 1 + \tan(x) $.
2. Find $ du = \sec^2(x) \, dx $.
3. Rewrite the integral in terms of $ u $: $ \int \frac{1}{u} \cdot \frac{1}{\sec^2(x)} \, du $.
4. Simplify the expression: $ \int \frac{\cos^2(x)}{u} \, du $.
5. Integrate $ \frac{1}{u} $ with respect to $ u $: $ \ln|u| + C $.
6. Substitute back $ u = 1 + \tan(x) $.

The final result is $ \ln|1 + \tan(x)| + C $, where $ C $ is the constant of integration.