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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

Sagot :

MrRoyal

Answer:

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c[/tex]

Step-by-step explanation:

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

[tex]\int {7 \sec(\theta) } \, d\theta[/tex]

Remove constant 7 out of the integrand

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) } \, d\theta[/tex]

Multiply by 1

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * 1} \, d\theta[/tex]

Express 1 as: [tex]\frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\sec(\theta) * \frac{\sec(\theta) + \tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta[/tex]

Expand

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{\sec^2(\theta) + \sec(\theta)\tan(\theta) }{\sec(\theta) + \tan(\theta)}} \, d\theta[/tex]

Let

[tex]u = \sec(\theta) + \tan(\theta)[/tex]

Differentiate

[tex]\frac{du}{d\theta} = \sec(\theta)\tan(\theta) + sec^2(\theta)[/tex]

Make [tex]d\theta[/tex] the subject

[tex]d\theta = \frac{du}{\sec(\theta)\tan(\theta) + sec^2(\theta)}[/tex]

So, we have:

[tex]\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)}[/tex]

Cancel out [tex]\sec(\theta)\tan(\theta) + sec^2(\theta)[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\int {\frac{1}{u}} \,du}}[/tex]

Integrate

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(u) + c[/tex]

Recall that: [tex]u = \sec(\theta) + \tan(\theta)[/tex]

[tex]\int {7 \sec(\theta) } \, d\theta = 7\ln(\sec(\theta) + \tan(\theta)) + c[/tex]

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