Question

In order to evaluate 7 sec(θ) dθ, multiply the integrand by sec(θ) + tan(θ) sec(θ) + tan(θ) . 7 sec(θ) dθ = 7 sec(θ) sec(θ) + tan(θ) sec(θ

Answer 1

Answer:

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$

Step-by-step explanation:

The question is not properly formatted. However, the integral of $\int {7 \sec(\theta) } \, d\theta$ is as follows:

$\int {7 \sec(\theta) } \, d\theta$

Remove constant 7 out of the integrand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta$

Multiply by 1

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta$

Express 1 as: $\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Expand

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta$

Let

$u = \sec(\theta) + \tan(\theta)$

Differentiate

$\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)$

Make $d\theta$ the subject

$d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

So, we have:

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{u}} \,* \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}$

Cancel out $\sec(\theta)\tan(\theta) + sec^2(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}$

Integrate

$\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c$

Recall that: $u = \sec(\theta) + \tan(\theta)$

$\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c$