# Thread: Integrating inverse of trig products

1. ## Integrating inverse of trig products

Hi
I'm kind of hunting out of my depth relative to my current knowledge so the answer to this might be really obvious - please bear with me ...

I'm trying to find
$\int \frac{1}{sin\theta cos\theta }d\theta$

I've tried integrating the inverse of a function
Inverse Function Integration -- from Wolfram MathWorld
but I ended up with an integral involving
$sin\left (sin\theta cos\theta \right )cos\left (sin\theta cos\theta \right )$
which didn't look very hopeful. It may be because I haven't officially learnt integrating an inverse yet ...

I then tried integration by parts (again something which I had never heard of before), and I changed the original integral to
$\int sec\theta csc\theta \, d\theta$
to do so, but then when it came to chosing which of sec or csc to integrate and which to differentiate, either way seemed to end up with the 'output' integral containing piles of ln, cot, csc, tan all in arrangements I couldn't hope to integrate.

Is there a technique(s) for integrating this which I don't know about yet? Any hints?

2. $\displaystyle \int\frac{1}{\sin{x}\cos{x}}\;{dx} = \int\frac{\cos^2{x}+\sin^2{x}}{\sin{x}\cos{x}}\;{d x}$ $\displaystyle = \int \frac{\cos{x}}{\sin{x}}\;d{x}-\int \frac{-\sin{x}}{\cos{x}}\;{dx} = \ln\left(\sin{x}\right)-\ln\left(\cos{x}\right)+k.$

3. An alternative:

$\displaystyle \int{\frac{1}{\sin{x}\cos{x}}\,dx} = \int{\frac{\cos{x}}{\sin{x}\cos^2{x}}\,dx}$

$\displaystyle = \int{\frac{\cos{x}}{\sin{x}(1 - \sin^2{x})}\,dx}$

Now make the substitution $\displaystyle u = \sin{x}$ so that $\displaystyle du = \cos{x}\,dx$ and the integral becomes

$\displaystyle \int{\frac{1}{u(1 - u^2)}\,du}$

which can be solved using Partal Fractions.

4. Ah ha! Thank you very much my friends

5. Hello, chug1!

Here's a sneaky approach . . .

$\displaystyle \int \frac{1}{\sin\theta\cos\theta}\,d\theta$

$\text{Divide numerator and denominator by }\cos^2\!\theta:\;\;\dfrac{\dfrac{1}{\cos^2\!\thet a}} {\dfrac{\sin\theta\cos\theta}{\cos^2\!\theta}} \;=\;\dfrac{\sec^2\!\theta}{\tan\theta}$

$\displaystyle \text{The integral becomes: }\;\int \frac{\sec^2\!\theta}{\tan\theta}\,d\theta \;=\;\ln|\tan\theta| + C$

6. Originally Posted by Soroban
Here's a sneaky approach . . .
Nice! I guess it's one of those integrals that can be evaluated in finitely many ways!

4. Letting $x = \arctan{t}$ gives:

\begin{aligned}\displaystyle \int\frac{1}{\sin{x}\cos{x}}\;{dx} & = \int\frac{2}{\sin{2x}}\;{dx} = \int\frac{2}{(1+t^2)\times \frac{2t}{1+t^2}}\;{dt} \\&= \int\frac{1}{t}\;{dt} = \ln{t}+k = \ln(\tan{x})+k.\end{aligned}

5. Letting $t = \frac{\sin{x}}{\cos{x}}$ gives:

\begin{aligned}\displaystyle \int\frac{1}{\sin{x}\cos{x}}\;{dx} & = \int\frac{\cos^2{x}}{\sin{x}\cos{x}}\;{dt} = \int\frac{\cos{x}}{\sin{x}}\;{dt} = \int\frac{1}{t}\;{dt} \\& = \ln{t}+k = \ln\left(\frac{\sin{x}}{\cos{x}}\right)+k = \ln\left(\tan{x}\right)+k.\end{aligned}

Notice that the above substitutions are actually equivalent!

7. An alternative :

$\displaystyle I = \int\frac{1}{sin(x)cos(x)} \, dx = 2 \int \dfrac{1}{sin(2x)} \, dx = 2 \int \, csc(2x) \, dx$

$u=2x \implies du=2dx \implies I = \int csc(u) \, du$ $= ln|csc(u)-cot(u)| + C = ln|csc(2x)-cot(2x)| + C$

8. Just a side note - what the OP posted are not inverse trigonometric functions, they are reciprocal trigonometric functions...

9. That's what I was about to say! (And will stick in my oar anyway!) It may be that the OP's problem was thinking that "inverse functions" ( $sin^{-1}= arcsin, cos^{-1}= arccos$) are the same as the reciprocals ( $\frac{1}{sin}= csc, \frac{1}{cos}= sec$) when he used "MathWorld" and so he entered the wrong functions. That's the problem with sites like that- they can't your thinking for you- GIGO!