Question

$\int \frac{1 dx}{sin(x-a)sin(x-b)}$

Answer 1

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

Answer 2

To solve the integral $\int \frac{1 dx}{\sin(x-a)\sin(x-b)}$, we use a standard technique for integrals of this form.

Let the integral be $I$. $I = \int \frac{1}{\sin(x-a)\sin(x-b)} dx$ We multiply the numerator and the denominator by $\sin(b-a)$, where $b-a$ is the difference between the two angles in the denominator. This is a constant provided $b \neq a + k\pi$ for any integer $k$.

$I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\sin(x-a)\sin(x-b)} dx$ We can rewrite the constant difference $b-a$ as the difference of the angles in the denominator: $(x-a) - (x-b) = x-a-x+b = b-a$. So, we substitute $\sin(b-a) = \sin((x-a) - (x-b))$ in the numerator.

$I = \frac{1}{\sin(b-a)} \int \frac{\sin((x-a) - (x-b))}{\sin(x-a)\sin(x-b)} dx$ Now, we use the trigonometric identity for the sine of a difference: $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Here, $A = x-a$ and $B = x-b$. $\sin((x-a) - (x-b)) = \sin(x-a)\cos(x-b) - \cos(x-a)\sin(x-b)$ Substitute this back into the integral: $I = \frac{1}{\sin(b-a)} \int \frac{\sin(x-a)\cos(x-b) - \cos(x-a)\sin(x-b)}{\sin(x-a)\sin(x-b)} dx$ Now, split the integrand into two separate fractions: $I = \frac{1}{\sin(b-a)} \int \left( \frac{\sin(x-a)\cos(x-b)}{\sin(x-a)\sin(x-b)} - \frac{\cos(x-a)\sin(x-b)}{\sin(x-a)\sin(x-b)} \right) dx$ Simplify each term: $I = \frac{1}{\sin(b-a)} \int \left( \frac{\cos(x-b)}{\sin(x-b)} - \frac{\cos(x-a)}{\sin(x-a)} \right) dx$ This simplifies to: $I = \frac{1}{\sin(b-a)} \int (\cot(x-b) - \cot(x-a)) dx$ We know that the integral of $\cot(u)$ with respect to $u$ is $\ln|\sin(u)|$. Applying this: $\int \cot(x-b) dx = \ln|\sin(x-b)|$ $\int \cot(x-a) dx = \ln|\sin(x-a)|$ So, the integral becomes: $I = \frac{1}{\sin(b-a)} (\ln|\sin(x-b)| - \ln|\sin(x-a)|) + C$ Using the logarithm property $\ln A - \ln B = \ln(A/B)$: $I = \frac{1}{\sin(b-a)} \ln\left|\frac{\sin(x-b)}{\sin(x-a)}\right| + C$ This solution is valid provided $\sin(b-a) \neq 0$, which means $b-a \neq k\pi$ for any integer $k$.

Explanation of the solution:

1. Multiply by $\sin(b-a)$: Introduce $\sin(b-a)$ in the numerator and denominator.
2. Rewrite Numerator: Express $\sin(b-a)$ as $\sin((x-a) - (x-b))$.
3. Expand: Use $\sin(A-B) = \sin A \cos B - \cos A \sin B$ to expand the numerator.
4. Split Fraction: Divide the numerator by the denominator $\sin(x-a)\sin(x-b)$ to get two terms.
5. Simplify: Cancel common factors to obtain $\cot(x-b) - \cot(x-a)$.
6. Integrate: Integrate each cotangent term: $\int \cot u \, du = \ln|\sin u|$.
7. Combine: Combine the results using logarithm properties.