Trigonometric anti derivative

Question

1. The question is the top integral ,I did substitution of 1+cos ,x+sin2x,sin,cos all of these not works, and I separate them out into two pieces and I am stuck in a new antiderivative :x /(1+cosx)²

\begin{align*} \newcommand{\dx}{\; \mathrm{d}x} \int \frac{x+\sin^2x}{(1+\cos x)^2} \dx &=\int \frac{x+(1-\cos^2x)}{(1+\cos x)^2} \dx \\ &=\int \frac{x}{(1+\cos x)^2} \dx + \int \frac{1-\cos^2x}{(1+\cos x)^2} \dx \\ &=\int \frac{x}{(1+\cos x)^2} \dx + \int \frac{(1-\cos x)(1+\cos x)}{(1+\cos x)^2} \dx \end{align*}

Answer 1

Using integration by parts gives $$ \int \dfrac{x}{(1 + \cos x)^2}dx = xv - \int v dx + C $$ where $v$ is an anti-derivative of $\displaystyle\int \dfrac{1}{(1 + \cos x)^2}dx$ and $C$ is a constant

Notice that, $1 + \cos x = 2\cos^2\left(\dfrac{x}{2}\right)$. Hence $$ \int \dfrac{1}{(1 + \cos x)^2}dx = \int \dfrac{dx}{4\cos^4 \frac{x}{2}} = \dfrac{1}{4}\int \left[1 + \tan^2 \left(\dfrac{x}{2}\right)\right]^2 dx $$ From here, let $t = \tan\left(\dfrac{x}{2}\right)$ and you'll be able to find $v$

Answer 2

The integrand is a product of an algebraic and a trigonometric function, so the ILATE mnemonic suggests integrating by parts, i.e. using the integration identity $$\int u\,dv = uv - \int v\,du ,$$ with $$u = x, \qquad dv = \frac{dx}{(1 + \cos x)^2} .$$

Integrating $dv$ gives

$$v = \int dv = \int \frac{dx}{(1 + \cos x)^2} .$$

One option (of many) for evaluating this latter integral is applying the so-called Weierstrass substitution, $$x = 2 \arctan t, \qquad dx = \frac{2\,dt}{1 + t^2} ,$$ which transforms the integral to $$\frac{1}{2} \int (t^2 + 1) \,dt .$$