Question

integral of sec(x+a) cosec(x+b)

Answer 1

The integral is:

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

Answer 2

Let the integral be $I = \int \sec(x+a) \operatorname{cosec}(x+b) dx$. We can rewrite the secant and cosecant functions in terms of sine and cosine: $I = \int \frac{1}{\cos(x+a)} \cdot \frac{1}{\sin(x+b)} dx = \int \frac{1}{\cos(x+a) \sin(x+b)} dx$.

To evaluate this integral, we use the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$. Let $A = x+b$ and $B = x+a$. Then $A-B = (x+b) - (x+a) = b-a$. We multiply the numerator and denominator by $\sin(b-a)$ (assuming $b-a \neq n\pi$ for integer $n$, so $\sin(b-a) \neq 0$): $I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\cos(x+a) \sin(x+b)} dx$ $I = \frac{1}{\sin(b-a)} \int \frac{\sin((x+b) - (x+a))}{\cos(x+a) \sin(x+b)} dx$ Using the sine subtraction formula in the numerator: $I = \frac{1}{\sin(b-a)} \int \frac{\sin(x+b) \cos(x+a) - \cos(x+b) \sin(x+a)}{\cos(x+a) \sin(x+b)} dx$

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

Simplify each term: $I = \frac{1}{\sin(b-a)} \int \left( \frac{\sin(x+b)}{\sin(x+b)} - \frac{\cos(x+b) \sin(x+a)}{\cos(x+a) \sin(x+b)} \right) dx$ $I = \frac{1}{\sin(b-a)} \int \left( 1 - \frac{\cos(x+b) \sin(x+a)}{\cos(x+a) \sin(x+b)} \right) dx$ $I = \frac{1}{\sin(b-a)} \int \left( 1 - \frac{\cos(x+b)}{\sin(x+b)} \cdot \frac{\sin(x+a)}{\cos(x+a)} \right) dx$ $I = \frac{1}{\sin(b-a)} \int \left( 1 - \cot(x+b) \tan(x+a) \right) dx$ $I = \frac{1}{\sin(b-a)} \int \left( 1 - \frac{\cos(x+b) \sin(x+a)}{\cos(x+a) \sin(x+b)} \right) dx$ $I = \frac{1}{\sin(b-a)} \int \left( \frac{\sin(x+b) \cos(x+a) - \cos(x+b) \sin(x+a)}{\sin(x+b) \cos(x+a)} \right) dx$ $I = \frac{1}{\sin(b-a)} \int \frac{\sin((x+b) - (x+a))}{\sin(x+b) \cos(x+a)} dx$ $I = \frac{1}{\sin(b-a)} \int \frac{\sin(b-a)}{\sin(x+b) \cos(x+a)} dx$ $I = \int \frac{1}{\sin(x+b) \cos(x+a)} dx$

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