Question

Integration of 1/sin(x-a)cos(x-b)

Answer 1

$\frac{1}{\cos(b-a)} \ln\left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C$

Answer 2

To integrate the expression $\frac{1}{\sin(x-a)\cos(x-b)}$, we use a trigonometric manipulation in the numerator.

Let the given integral be $I$. $I = \int \frac{1}{\sin(x-a)\cos(x-b)} dx$

We observe the arguments of the trigonometric functions are $(x-a)$ and $(x-b)$. Their difference is $(x-a) - (x-b) = b-a$. We can introduce $\cos(b-a)$ in the numerator, as it is a constant with respect to $x$. To do this, we multiply and divide by $\cos(b-a)$. $I = \frac{1}{\cos(b-a)} \int \frac{\cos(b-a)}{\sin(x-a)\cos(x-b)} dx$

Now, we use the identity $b-a = (x-a) - (x-b)$. Let $P = x-a$ and $Q = x-b$. The numerator becomes $\cos(P-Q)$. Using the trigonometric identity $\cos(P-Q) = \cos P \cos Q + \sin P \sin Q$: $\cos((x-a) - (x-b)) = \cos(x-a)\cos(x-b) + \sin(x-a)\sin(x-b)$

Substitute this into the integral: $I = \frac{1}{\cos(b-a)} \int \frac{\cos(x-a)\cos(x-b) + \sin(x-a)\sin(x-b)}{\sin(x-a)\cos(x-b)} dx$

Now, split the fraction into two terms: $I = \frac{1}{\cos(b-a)} \int \left( \frac{\cos(x-a)\cos(x-b)}{\sin(x-a)\cos(x-b)} + \frac{\sin(x-a)\sin(x-b)}{\sin(x-a)\cos(x-b)} \right) dx$

Simplify each term: The first term simplifies to $\frac{\cos(x-a)}{\sin(x-a)} = \cot(x-a)$. The second term simplifies to $\frac{\sin(x-b)}{\cos(x-b)} = \tan(x-b)$.

So the integral becomes: $I = \frac{1}{\cos(b-a)} \int (\cot(x-a) + \tan(x-b)) dx$

Now, integrate term by term using the standard integral formulas: $\int \cot u du = \ln|\sin u| + C$ $\int \tan u du = -\ln|\cos u| + C$

$I = \frac{1}{\cos(b-a)} [\ln|\sin(x-a)| - \ln|\cos(x-b)|] + C$

Using the logarithm property $\ln M - \ln N = \ln\left(\frac{M}{N}\right)$: $I = \frac{1}{\cos(b-a)} \ln\left|\frac{\sin(x-a)}{\cos(x-b)}\right| + C$

This solution is valid provided $\cos(b-a) \neq 0$. If $\cos(b-a) = 0$, then $b-a = (2n+1)\frac{\pi}{2}$ for some integer $n$, which leads to a special case where the denominator simplifies differently (e.g., to $\sin^2(x-a)$ or $\cos^2(x-a)$).

Explanation of the solution: The integral is solved by introducing a constant trigonometric term, $\cos(b-a)$, in the numerator. This term is then expanded using the angle subtraction formula for cosine, $\cos(P-Q) = \cos P \cos Q + \sin P \sin Q$, where $P=x-a$ and $Q=x-b$. The expanded numerator is then divided by the original denominator, which results in a sum of $\cot(x-a)$ and $\tan(x-b)$ terms. These terms are then integrated separately using standard integration formulas for cotangent and tangent, and the results are combined using logarithm properties.