Question

$$ \int e^x (\tan x + \sec^2 x) \, dx = ? $$

• A. $e^x \sec x + c$
• B. $e^x \tan x + c$
• C. $e^x \cot x + c$
• D. $e^x \tan^2 x + c$

Hint:

Always check the derivative of the first term inside the bracket. In trigonometric forms involving $e^x$, common pairs are $(\sin x, \cos x)$, $(\tan x, \sec^2 x)$, and $(\sec x, \sec x \tan x)$.

Answer 1

The Correct Option is B

To solve the integral \(\int e^x (\tan x + \sec^2 x) \, dx\), let's break it down step by step:

1. Observe the integral's form: \(\int e^x (\tan x + \sec^2 x) \, dx\). We can split it into two separate integrals: \(\int e^x \tan x \, dx + \int e^x \sec^2 x \, dx\).
2. Begin by solving each integral separately. Let's start with \(\int e^x \tan x \, dx\). This integral can be attacked through integration by parts, where we let:
   • \(u = \tan x\)and \(dv = e^x \, dx\).
   • Differentiating and integrating:
     • \(du = \sec^2 x \, dx\)
     • \(v = e^x\)
   • Using the integration by parts formula, \(\int u \, dv = uv - \int v \, du\):
     • \(\int e^x \tan x \, dx = e^x \tan x - \int e^x \sec^2 x \, dx\)
3. Now, handle the second integral \(\int e^x \sec^2 x \, dx\). This form is already present in our existing equation from the integration by parts step:

Notice from our integration by parts equation that we get back \(-\int e^x \sec^2 x \, dx\)which cancels itself out, leaving us with the term:

• \(e^x \tan x\).

Thus, the final solution to the integral \(\int e^x (\tan x + \sec^2 x) \, dx\)is:

• Answer: \(e^x \tan x + c\), where \(c\)is the constant of integration.