Integrate (tan x +tan^3 x)/(1+tan^3 x)?
1 Answer
Explanation:
We can rewrite the numerator as:
#tanx(1 + tan^2x)#
We know that
#I = int(tanxsec^2x)/(1 + tan^3x)dx#
Let
#I = int (tanxsec^2x)/(1 + tan^3x) * (du)/sec^2x#
#I = int tanx/(1 + tan^3x) du#
Now we need to change into
#I =int u/(1 + u^3) du#
We now observe that
#I = int u/((u + 1)(u^2 - u + 1)) du#
Now by partial fractions, we have:
#A/(u + 1) + (Bu+ C)/(u^2 - u + 1) = u/((u + 1)(u^2 - u + 1))#
#A(u^2 - u + 1) + (Bu + C)(u + 1) = u#
#Au^2 - Au + A + Bu^2 + Cu + B u + C = u#
Now write a system of equations in 3 variables.
#{(A + B = 0), (B + C - A = 1), (A + C = 0):}#
We can immediately see that
#I = int -1/(3(u + 1)) + (1/3u + 1/3)/(u^2 - u + 1)du#
#I = -1/3ln|u + 1| + 1/3int (u + 1)/(u^2 - u + 1) du#
Now we have another integral to solve. Call it
#I_2 = 1/3int (u + 1)/(u^2 - u + 1)#
Now we can rewrite as
#I_2 = 1/3int (1/2(2u - 1) + 3/2)/(u^2 - u + 1) du#
We can now separate into two integrals.
#I_2 = 1/3int(1/2(2u - 1))/(u^2- u + 1) + 3/(2(u^2 - u + 1))#
#I_2 = 1/6 int (2u -1)/(u^2 - u + 1) + 3/(2(u^2 - u + 1)#
Now we let
#I_2 = 1/6int (2u - 1)/(u^2 - u + 1) * (dn)/(2u - 1) +int 3/(2(u^2 - u + 1))du#
#I_2 = 1/6int 1/n dn + int 3/(2(u^2 - u + 1))du#
#I_2 = 1/6ln|n| + int 3/(2(u^2 - u + 1))du#
We now have a third integral that needs to be solved. Call it
#I_3 = 3/2int 1/(1(u^2 - u + 1/4 - 1/4) + 1) du#
#I_3 = 3/2int 1/((u - 1/2)^2 + 3/4) du#
#I = 3/2int 1/(((2u - 1)/2)^2 + 3/4)du#
We now let
#I_3 =(3)/(4sqrt(3))int 1/((sqrt(3)/2t)^2 + 3/4) dt#
#I_3= 3/(4sqrt(3))int 1/(3/4(t^2 + 1)) dt#
#I_3 = 1/sqrt(3)int 1/(t^2 + 1) dt#
#I_3 = 1/sqrt(3)arctan(t)#
We now must reverse the substitutions.
#I_3 = sqrt(3)arctan((2u - 1)/sqrt(3))#
So putting everything back together:
#I = 1/6ln|u^2 - u + 1| + 1/sqrt(3)arctan((2u - 1)/sqrt(3)) - 1/3ln|u + 1| + C#
#I = 1/6ln|tan^2x - tanx + 1| + 1/sqrt(3)arctan((2tanx - 1)/sqrt(3)) - 1/3ln|tanx+ 1| +C#
Hopefully this helps!