What is #int tan^3(x) sec^4(x/2) dx#?

1 Answer
Jun 17, 2016

#8/(1-tan^2(x/2))^2-24/(1-tan^2(x/2))-8ln(abs(1-tan^2(x/2)))+C#

Explanation:

The first step here should be to eliminate the half-angle--even though it will crate a double angle in the tangent function, double-angles are easier to work with.

Let #u=x/2#, such that #x=2u# and #du=1/2dx# and #dx=2du#.

Substituting these in, we see that:

#inttan^3(x)sec^4(x/2)dx=2inttan^3(2u)sec^4(u)du#

Note the tangent double-angle formula:

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

Therefore:

#2inttan^3(2u)sec^4(u)du=2intsec^4(u)[(2tan(u))/(1-tan^2(u))]^3du#

Now, cube the fraction and take a #sec^2(u)# from the #sec^4(u)# and write it as #1+tan^2(u)#. This gives us an integrand of entirely tangents, except for a #sec^2(u)#, its derivative.

#=2intsec^2(u)((8tan^3(u)(1+tan^2(u)))/(1-tan^2(u))^3)du#

Now, let #v=tan(u)# such that #dv=sec^2(u)du#.

#=16int(v^3(1+v^2))/(1-v^2)^3dv#

Now, let #w=1-v^2# so that #dw=-2vdv#. Rearranging so that #-2v# is present:

#=-8int(-2v(v^2)(1+v^2))/(1-v^2)^3dv#

Here, from #w=1-v^2#, we see that #v^2=1-w# and #1+v^2=2-w#. Hence:

#=-8int((1-w)(2-w))/w^3dw#

Expanding:

#=-8int(2-3w+w^2)/w^3dw#

Splitting and writing with negative exponents, except for on the last one (since it's the natural logarithm integral:

#=-16intw^-3dw+24intw^-2-8int1/wdw#

Integrating using #intw^ndw=w^(n+1)/(n+1)+C# and #int1/wdw=ln(absw)+C#, we see that:

#=-16(w^-2/(-2))+24(w^-1/(-1))-8ln(absw)+C#

Using #w=1-v^2#:

#=8/(1-v^2)^2-24/(1-v^2)-8ln(abs(1-v^2))+C#

Using #v=tan(u)#:

#=8/(1-tan^2(u))^2-24/(1-tan^2(u))-8ln(abs(1-tan^2(u)))+C#

Using #u=x/2#:

#=8/(1-tan^2(x/2))^2-24/(1-tan^2(x/2))-8ln(abs(1-tan^2(x/2)))+C#