# Trigonometric anti derivative

calculusindefinite-integralsintegration

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 ðŸ˜¡ /(1+cosx)Â²

\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*}

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 $$v$$