Integration of $\int \frac{(1 + x)\sin x}{(x^2 +2 x)\cos^2 x-(1 + x)\sin2x}dx$

Question

The integral is $$\int \dfrac{(1 + x)\sin x}{(x^2 + 2x)\cos^2 x-(1 + x)\sin2x}dx.$$I've tried the problem by first multiplying both the numerator and denominator by $\sec^2 x$ but couldn't do justice. Can anyone help, please?

Answer 1

Let $$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x-(1+x)\sin 2x}dx$$

$$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x+1)\cos^2 x-(1+x)\sin 2x-\cos^2 x}dx$$

$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+2)\sin x\cdot \cos x-(1-\sin^2 x)}dx$$

So $$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+1)\sin x\cdot \cos x+\sin^2 x-1}dx$$

$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x-\sin x\right]^2-1^2}dx$$

Now Let $(x+1)\cos x-\sin x = t\;,$ Then $(x+1)\sin xdx = -dt$

So Integral $$\displaystyle I = -\int\frac{1}{t^2-1}dt = -\frac{1}{2}\int\left[\frac{1}{t-1}-\frac{1}{t+1}\right]dt$$

So we get $$\displaystyle I = \frac{1}{2}\left[\ln|t+1|-\ln|t-1|\right]+\mathcal{C} = \frac{1}{2}\ln\left|\frac{t+1}{t-1}\right|+\mathcal{C}$$

So we get $$\displaystyle I = \frac{1}{2}\ln \left|\frac{(x+1)\cos x-\sin x+1}{(x+1)\cos x-\sin x-1}\right|+\mathcal{C}$$

Answer 2

It might help to factorize the denominator and do partial fraction this way

Denominator$=(x^2+2x+1)\cos^2(x)-(1+x)\sin2x-\cos^2x\\=\cos^2x[(x+1)^2-2(x+1)\tan(x)-1]\\=\cos^2x(x+1-(\tan(x)+\sec(x)))(x+1-(\tan(x)-\sec(x)))$

The answer without process can be found using http://www.wolframalpha.com, which involves lots of natural log function, which is why I think partial fraction is the way to do.