Evaluate: $\int \frac {\sec (x)}{\sec (x)+\tan (x)} dx$

Question

Evaluate: $\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx$

My Attempt: $$=\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx$$ $$=\int \dfrac {\sec^2 (x)}{\sec^2 (x)+\sec (x).\tan (x)}$$

Answer 1

$$\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx =\int \sec (x)(\sec x - \tan x) dx = \int \sec^2 x -\tan x \sec x dx \\= \tan x - \sec x + C$$

Answer 2

Or...if you are not too keen on working with trig reciprocal functions: Multiply top and bottom by $\cos x$ to arrive at $\int{\frac{dx}{1+\sin x}}$. Then multiply top and bottom by $1-\sin x$ to arrive at $\int{\frac{1-\sin x}{\cos^2 x}}dx$. Now split the fraction and it becomes standard work. It may not be quicker but at least you are working with the more familiar $\sin/\cos$ functions...

Answer 3

Another approach might be the following:

$$\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx =\int \dfrac {\sec (x)(\sec (x)+\tan (x))}{(\sec (x)+\tan (x))^2} dx \to\begin{bmatrix}u=\sec (x)+\tan (x)\\ du=\sec(x)(\sec (x)+\tan (x))\,dx\end{bmatrix}\to\int\frac1{u^2}\,du=\boxed{-\frac1{\sec(x)+\tan(x)}+C}$$

This method works for any integral of the form $$\int \dfrac {\sec (x)}{(\sec (x)+\tan (x))^n} dx $$ with $n\in \mathbb{Z}$.

Answer 4

Another approach $$ \begin{align} I&=\int \dfrac {\sec (x)}{\sec (x)+\tan (x)} dx\\ &=\int \dfrac {dx }{1+\sin (x)} \\ &=\int \dfrac {dx}{\sin(\pi/2)+\sin (x)} \\ &=\int \dfrac {dx}{2\cos(\pi/4 -x/2)\sin (x/2+\pi/4)} \\ &=\int \dfrac {dx}{2\sin^2 (x/2+\pi/4)} \\ I&=- \cot(x/2+\pi/4) +K \end{align} $$

Answer 5

Another approach:

$$\begin{align}\int\ \dfrac{\sec x}{\sec x+\tan x}\,\mathrm dx &=\int\ \dfrac{\sec^2 x-\sec x\cdot\tan x}{\sec^2 x-\tan^2 x}\,\mathrm dx \\&=\int \sec^2 x\,\mathrm dx-\int\ \sec x\cdot \tan x\,\mathrm dx\\ &=\boxed{\tan x-\sec x+C}\end{align}$$