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Find $\displaystyle \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx$

I couldn't find my answer so I looked up the solution which is as follows $$\displaystyle \begin{align}&\int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}\,dx\\&= \int \frac{2\cos x+1}{(2+\cos x)^2}\,dx\\&= \int \frac{2\cos x+\cos^2 x+\sin^2 x}{(2+\cos x)^2}\,dx\\&=\int \frac{\cos x}{2+\cos x}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{1}{2+\cos x}\cdot(\sin x)-\int \sin x\cdot\frac{\sin x}{(2+\cos x)^2}\,dx+\int \frac{\sin^2 x}{(2+\cos x)^2}\,dx\\&=\frac{\sin x}{2+\cos x}+C\end{align}$$ I was stuck with $\displaystyle\int \frac{2\cos x+1}{(2+\cos x)^2}\,dx$ and didn't think of replacing $1$ with $\cos^2 x+\sin^2 x$. I want to know the motivation behind this because the reason of this substitution wasn't very obvious to me at first and it only became clear to me in the second last step. Other ways to solve this are also welcome.

VIVID
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Asher2211
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    This method is non-obvious, and I doubt many people would think to solve the integral this way. However, any function that can be written as the quotient of two trigonometric expressions can be integrated using the substitution $t=\tan(x/2)$. Note that the substitution only works if the only terms that appear in the integrand are $\sin x$ and $\cos x$. – Joe Jun 05 '21 at 20:30
  • @Joe Does the substitution by $\tan \frac{x}{2}$ applicable in all cases where the only terms that appear in the integrand are $\sin x$ and $\cos x$? – Asher2211 Jun 05 '21 at 20:59
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    Yes, it works in all cases, hence why it is such a useful method. The worst case scenario is that you get some ugly partial fractions, but that's better than not being able to solve the integral at all. – Joe Jun 05 '21 at 21:28

2 Answers2

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Using the "universal substitution": $$\begin{align} \int \frac{\sec x(2+\sec x)}{(1+2\sec x)^2}dx &= \int\frac{(\tan^2(x/2) + 1)\left(2+ \frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)}{(1-\tan^2(x/2))\left(1+\frac{\tan^2(x/2) + 1}{1-\tan^2(x/2)}\right)^2} \\ &= \begin{bmatrix}u = \tan(x/2) \\ du = \frac{1}{2}\sec^2(x/2)dx\end{bmatrix} \\ &= -2 \int \frac{u^2-3}{(u^2 + 3)^2}du,\end{align}$$ which can be integrated using the known algorithms. However, the original solution is indeed more attractive.

amWhy
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VIVID
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Analogous of how integration by parts comes from the product rule, from the quotient rule of differentiation:

$$\begin{align*} \frac uv &= \int \frac {du}{v} - \int\frac{u\ dv}{v^2}\\ \int\frac{u\ dv}{v^2} &= -\frac uv + \int \frac {du}{v}\\ \end{align*}$$

Or directly deriving from integration by parts:

$$ \int \frac{u\ dv}{v^2} = \int u\ d\left(-\frac1v\right) = -\frac{u}{v} + \int\frac{du}{v} $$

This might have motivated breaking down $1$ into $\cos^2 x + \sin^2 x$, to apply the quotient rule backwards,

$$\begin{align*} I &= \int \frac{\cos x}{2+\cos x} dx + \int \frac{\sin^2 x}{(2+\cos x)^2}dx\\ &= \int\frac{d(\sin x)}{2+\cos x} - \int \frac{\sin x\ d(2+\cos x)}{(2+\cos x)^2} \end{align*}$$

But the exact break down of the numerator was not obvious to me. Though by this special case of integration by parts via the quotient rule, let

$$\begin{align*} v &= 2 + \cos x & dv &= -\sin x\ dx\\ u &= -(2\cot x +\csc x)& du &= (2\csc^2x + \csc x \cot x) dx \end{align*}$$

$$\begin{align*} \int \frac{2\cos x + 1}{(2+\cos x)^2} dx &= \int\frac{-(2\cot x + \csc x)(-\sin x\ dx)}{(2+\cos x)^2}\\ &= \frac{2\cot x + \csc x}{2+\cos x} + \int \frac{2\csc^2 x + \csc x \cot x}{2+\cos x} dx\\ &= \frac{2\cot x + \csc x}{2+\cos x} + \int \csc^2 x\cdot \frac{2+\cos x}{2+\cos x} dx\\ &= \frac{2\cot x + \csc x}{2+\cos x} + \int \csc^2 x\ dx\\ &= \frac{2\cot x + \csc x}{2+\cos x} - \cot x + C\\ &= \frac{(2\cot x + \csc x) - \cot x(2+\cos x)}{2+\cos x} + C\\ &= \frac{\csc x - \cot x\cos x}{2+\cos x} + C\\ &= \frac{\sin x (\csc^2 x - \cot^2 x)}{2+\cos x} + C\\ &= \frac{\sin x}{2+\cos x} + C\\ \end{align*}$$

Alternatively, without multiplying the starting numerator and denominator by $\cos^2 x$, let

$$\begin{align*} v &= 1 + 2\sec x & dv &= 2\sec x \tan x\ dx\\ u &= \frac12(2\cot x + \csc x)& du &= -\frac12(2\csc^2x + \csc x \cot x) dx \end{align*}$$

$$\begin{align*} \int \frac{\sec x(2+\sec x)}{(1+ 2\sec x)^2} dx &= \int\frac{\frac12(2\cot x + \csc x)(2\sec x \tan x\ dx)}{(1+2\sec x)^2}\\ &= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} + \int\frac{-\frac12 (2\csc^2 x + \csc x \cot x)}{1+2\sec x} dx\\ &= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \int(-\csc x \cot x)\cdot\frac{2\sec x + 1}{1+2\sec x} dx\\ &= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \int(-\csc x \cot x)\ dx\\ &= -\frac{\frac12(2\cot x+\csc x)}{1+2\sec x} +\frac12 \csc x + C\\ &= \frac{-\cot x-\frac12\csc x + \frac12\csc x(1+2\sec x)}{1+2\sec x} + C\\ &= \frac{-\cot x+ \csc x\sec x}{1+2\sec x} + C\\ &= \frac{\tan x(-\cot^2 x+ \csc^2 x)}{1+2\sec x} + C\\ &= \frac{\tan x}{1+2\sec x} + C\\ \end{align*}$$

peterwhy
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