# Integrate 1/ ( sin x + sec x)

• velliyangiri
So in summary, try substituting:\begin{align*}\frac{1}{\sin x + \sec x} &= \frac{1}{\sin x + \sec x} \cdot \frac{\sin x - \sec x}{\sin x - \sec x} \\&= \frac{\sin x - \sec x}{\sin^2 x - \sec^2 x} \\&= \frac{\cos^2 x\sin x - \cos x}{\cos^2 x\ \sin^2 x - 1} \\&= \frac{\cos^2 x}{\cos^2 x\ (1-\cos^2

## Homework Statement

∫ 1 /( sin x + sec x) dx

## The Attempt at a Solution

∫ cos x / ( sin x + cos x ) style question
Tried ∫ cos x / (sin x . cos x + 1) dx
and uses sin 2x
tried substitutions
nothing seem to work

I'm really curious to see if anyone can get this one. I plotted the graph and it looks very strange.. Please post the solution when you find it, it would help me sleep at night.

Highway said:

Lol.. Oh my..

There has to be an easier way to get this one.

You can always reduce it to an integral with rational algebraic expressions with the substitution:
$$t \equiv \tan \left( \frac{x}{2} \right)$$

$$\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}$$

Dickfore said:
You can always reduce it to an integral with rational algebraic expressions with the substitution:
$$t \equiv \tan \left( \frac{x}{2} \right)$$

$$\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}$$

man, i wish i knew how to do this back in the day. . .

Dickfore said:
You can always reduce it to an integral with rational algebraic expressions with the substitution:
$$t \equiv \tan \left( \frac{x}{2} \right)$$

$$\sin x = \frac{2 t}{1 + t^2}, \ \cos x = \frac{1 - t^2}{1 + t^2}, \ dx = \frac{2 \, dt}{1 + t^2}$$

I can't see that being much easier because after playing with the algebra it's going to leave many t's with various higher degrees (up to 6th degree if my algebra is correct) and also in a fraction form. After plugging back in tan it's going to have some subs and then reduction formulas for those higher degrees.

and this is exactly why the output from Wolfram Alpha looks that long. The only difference is that the imaginary unit is missing and instead of inverse hyperbolic, you'll have inverse trigonometric functions.

No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.

You could try something like this:
\begin{align*}
\frac{1}{\sin x + \sec x} &= \frac{1}{\sin x + \sec x} \cdot \frac{\sin x - \sec x}{\sin x - \sec x} \\
&= \frac{\sin x - \sec x}{\sin^2 x - \sec^2 x} \\
&= \frac{\cos^2 x\sin x - \cos x}{\cos^2 x\ \sin^2 x - 1} \\
&= \frac{\cos^2 x}{\cos^2 x\ (1-\cos^2 x) - 1}\sin x - \frac{1}{(1-\sin^2 x)\sin^2 x - 1}\cos x
\end{align*}Then use a trig substitution for each of the two terms.

Dickfore said:
No one said the result should have a simple form. Given the integrand, no other simpler substitution seems possible. But, you need to do the partial fraction decomposition yourself, or at least bring the integral in a rational algebraic form if you want further help.

Yeah I know what you mean, it's just that the integral just looked deceiving simple at first glance.

DrummingAtom said:
Lol.. Oh my..

There has to be an easier way to get this one.
There is.

Use the fact that cos(-pi/4) = sin(-pi/4)=1/√2 .

\begin{aligned} \sin x + \cos x &= \sqrt2 \left(\cos x\frac 1 {\sqrt2} + \sin x \frac 1 {\sqrt2}\right) \\ &= \sqrt2 (\cos x \cos(\pi/4) + \sin x \sin (\pi/4)) \\ &= \sqrt2 \cos(x-\pi/4) \end{aligned}

And thus $1/(\sin x + \cos x) = \sec(x-\pi/4)/\sqrt 2$

So now all you have to do is find the integral of sec(x-pi/4) and then divide by √2.

the only problem is that the denominator is $\sin x + \sec x$, and not $\sin x + \cos x$ as you were trying to simplify.