Integral $\int\frac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$

Question

Integrate $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$

I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.

Answer 1

After substituting $\cos(2x)=\cos^2x-\sin^2x$ and $\sin(2x)=2\sin x\cos x$ and doing a bit of algebra, we find that

$$2x\cos(2x)+(x^2-1)\sin(2x)=2(x\sin x+\cos x)(x\cos x-\sin x)$$

and thus

$$\begin{align}{2x^2\over 2x\cos(2x)+(x^2-1)\sin(2x)} &={x^2\over(x\sin x+\cos x)(x\cos x-\sin x)}\\ &={x\cos x\over x\sin x+\cos x}+{x\sin x\over x\cos x-\sin x}\end{align}$$

The substitution $u=x\sin x+\cos x$, $du=x\cos x\,dx$ tells us

$$\int{x\cos x\over x\sin x+\cos x}\,dx=\int{du\over u}=\ln|u|+C=\ln|x\sin x+\cos x|+C$$

while the substitution $u=x\cos x-\sin x$, $du=-x\sin x$ tells us

$$\int{x\sin x\over x\cos x-\sin x}\,dx=-\int{du\over u}=-\ln|u|+C = -\ln|x\cos x-\sin x|+C$$

and thus

$$\int{2x^2\over 2x\cos(2x)+(x^2-1)\sin(2x)}\,dx=\ln\left|x\sin x+\cos x\over x\cos x-\sin x\right|+C$$

Remark: The key here was to get split the integral into two pieces, each of which is easily handled with a simple substitution. The two pieces' substitutions are similar, but not the same. It was not obvious (to me, at least) that the double-angle formulae would lead to a nice factorization, nor that "partial fractions" on the factored denominator would produce such nice results, but the individual steps seemed natural to consider.

Answer 2

The hint:

Compute $$\left(\ln\dfrac{x-\tan x}{x\tan x+1}\right)^{'}.$$

Answer 3

$$2x\cos (2x)+(x^2-1)\sin 2x=(x^2+1)\bigg[\frac{2x}{x^2+1}\cos 2x+\frac{x^2-1}{x^2+1}\sin 2x\bigg]$$

$$=(x^2+1)\cos\bigg(2x-2\alpha\bigg)$$

$\displaystyle \sin(2\alpha)=\frac{x^2-1}{x^2+1}$ and $\displaystyle \cos(2\alpha)=\frac{2x}{x^2+1}$

integration $$=\int\sec\bigg(2x-\tan^{-1}\frac{x^2-1}{2x}\bigg)\frac{2x^2}{x^2+1}dx$$

put $\displaystyle 2x-\tan^{-1}\bigg(\frac{x^2-1}{2x}\bigg)=t$

And $\displaystyle \frac{2x^2}{x^2+1}dx=dt$

Integral $$=\int \sec(t)dt=\ln\bigg|\sec (t)+\tan (t)\bigg|+C$$

Answer 4

Let $\arctan x=y\implies x=\tan y,x^2-1=\dfrac{\sin^2y}{\cos^2y}-1=-(1+x^2)\cos2y$

$$2x\cos(2x)+(x^2-1)\sin2x=(1+x^2)\sin2(\arctan x-x)$$

$$I=\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} =\int\dfrac{2x^2}{(1+x^2)\sin2(\arctan x-x)}$$

Now set $u=\arctan x-x,du=-\dfrac{x^2\ dx}{1+x^2}$

$$I=-2\int\dfrac{du}{\sin2u}=?$$

Use this

Answer 5

Here's a solution using @MichaelRozenberg's hint. Solution requires some careful observation and without Michael's hint, I wouldn't have noticed it. Still, it provides a quick answer.

Using double-angle formula, we get

$$\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)}=\dfrac{x^2}{(x\cos x-\sin x)(x\sin x+\cos x)}.$$ Dividing the numerator and denominator by $\cos^2x $ gives, $$\dfrac{x^2\sec^2x}{(x-\tan x)(x\tan x+1)}.$$ Now observe that $$\left(\dfrac{x-\tan x}{x\tan x+1}\right)^{'}=\dfrac{-x^2\sec^2x}{(x\tan x+1)^2}.$$ So multiply the numerator and denominator by $x\tan x+1$ and take $\dfrac{x-\tan x}{x\tan x+1}=t. $