Question

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

Answer 1

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

Answer 2

To solve the integral $\int \sec(x+a)\sec(x+b) dx$, we can rewrite the secant functions in terms of cosine:

$I = \int \frac{1}{\cos(x+a)\cos(x+b)} dx$

We can use a trigonometric identity to manipulate the numerator. Consider the term $\sin(a-b)$. We can write this as $\sin((x+a)-(x+b))$. Using the sine subtraction formula, $\sin(A-B) = \sin A \cos B - \cos A \sin B$, we have:

$\sin(a-b) = \sin((x+a)-(x+b)) = \sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)$

We can multiply and divide the integrand by $\sin(a-b)$, assuming $\sin(a-b) \neq 0$ (i.e., $a-b \neq n\pi$ for any integer $n$).

$I = \frac{1}{\sin(a-b)} \int \frac{\sin(a-b)}{\cos(x+a)\cos(x+b)} dx$

Substitute the expanded form of $\sin(a-b)$ in the numerator:

$I = \frac{1}{\sin(a-b)} \int \frac{\sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)} dx$

Now, split the fraction into two terms:

$I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x+a)\cos(x+b)}{\cos(x+a)\cos(x+b)} - \frac{\cos(x+a)\sin(x+b)}{\cos(x+a)\cos(x+b)} \right) dx$

Simplify the terms:

$I = \frac{1}{\sin(a-b)} \int \left( \frac{\sin(x+a)}{\cos(x+a)} - \frac{\sin(x+b)}{\cos(x+b)} \right) dx$

$I = \frac{1}{\sin(a-b)} \int (\tan(x+a) - \tan(x+b)) dx$

Now, integrate each term. The integral of $\tan(u)$ is $-\ln|\cos(u)| + C$.

$\int \tan(x+a) dx = -\ln|\cos(x+a)| + C_1$

$\int \tan(x+b) dx = -\ln|\cos(x+b)| + C_2$

Substitute these back into the expression for $I$:

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

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

Using the logarithm property $\ln A - \ln B = \ln(A/B)$:

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

Alternatively, using the integral $\int \tan(u) du = \ln|\sec(u)| + C$:

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

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

Since $\sec(u) = 1/\cos(u)$, $\frac{\sec(x+a)}{\sec(x+b)} = \frac{1/\cos(x+a)}{1/\cos(x+b)} = \frac{\cos(x+b)}{\cos(x+a)}$, so both forms are equivalent.

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