$\int \cos^4{x}dx$ unsolvable with $t = \tan{x}$?

Question

I have been told that "universal substitution always works", so I wanted to give it a try on this specific integral.

$\int \cos^4{x}dx$

For some reason it does not work. Please note that I solved this integral in a normal way, just don't understand why the universal substitution does not work.

$\Bigg(t=\tan{x}, \cos^2{x} = \frac{1}{1+t^2},dx=\frac{dt}{1+t^2} \Bigg)$

$\int \cos^4xdx = \int (\cos^2x)^2dx = \int \Big(\frac{1}{1+t^2} \Big)^2 \frac{dt}{1+t^2} = \int \frac{dt}{(1+t^2)(1+t^2)(1+t^2)}$

$\frac{1}{(1+t^2)^3} = \frac{At +B}{1+t^2} + \frac{Ct+D}{(1+t^2)^2} + \frac{Et+F}{(1+t^2)^3}$

$1 = (At+B)(t^4+2t^2+1)+(Ct+D)(1+t^2)+Et+F$

$1 = At^5 + 2At^3 + At + Bt^4 + 2Bt^2 + B + Ct^3 + Ct + Dt^2 + D + Et + F$

Now this boils down to six linear equations:

$0 = A$

$0 = B$

$0 = 2A + C$

$0 = 2B + D$

$0 = A + C + E$

$1 = B + D + F$

Which results in: $A = 0, B = 0, C = 0, D = 0, E = 0, F = 0$ making it unsolvable for me.

Any ideas what went wrong?

Answer 1

You made a mistake in solving your system of equations. The solution is $$A = 0, B = 0, C = 0, D = 0, E = 0, F = 1$$

You could have seen this directly by observing that $\frac{1}{(1+t^2)^3}$ is already in the desired form: $$\frac{1}{(1+t^2)^3} = \frac{0t +0}{1+t^2} + \frac{0t+0}{(1+t^2)^2} + \frac{0t+1}{(1+t^2)^3}$$

You could have skipped this step and go directly to the "reduction formula" $$\int\frac{dt}{(t^2+1)^m}=\frac{t}{2(m-1)(t^2+1)^{m-1}}+\frac{2m-3}{2m-2}\int\frac{dt}{(t^2+1)^{m-1}}$$ (or integration by parts if you are not familiar with this).

Answer 2

Your attempt at partial fraction decomposition is the problem; you can't remedy repeated factors like that. (Well, not usefully; see my comment.)

Instead you could cube $\frac{1}{1+t^2}=\frac{1}{2i}\left(\frac{1}{t-i}-\frac{1}{t+i}\right)$ and simplify further. I don't recommend it, though. The simplest solution for the original problem is $$\int\cos^4 xdx=\frac{1}{4}\int(1+\cos 2x)^2dx=\frac{1}{8}\int(3+4\cos 2x+\cos 4x)dx=\frac{3x+2\sin 2x+\frac{1}{4}\sin 4x}{8}+C.$$

Answer 3

Another way using reduction formula

$$\dfrac{d(\cos^nx\sin x}{dx}=\cos^{n+1}x-n\cos^{n-1}x(\cos^2x-1)=-(n-1)\cos^{n+1}x+n\cos^{n-1}x$$

Integrate both sides $$\cos^nx\sin x+K=-(n-1)I_{n+1}+nI_{n-1}$$

where $I_m=\int\cos^mx\ dx$ and $k$ is an arbitrary constant

Set $n=3,1$