help with part b ! - please see attachment

Set u:
[tex]\displaystyle u = \sec x[/tex]
[u] Apply Trigonometric Differentiation:
[tex]\displaystyle du = \sec x \tan x \ dx[/tex]
[du] Rewrite [U-Solve]:
[tex]\displaystyle dx = \cos x \cot x \ du[/tex]

Step 3: Integrate Pt. 2

[Integral] Apply Integration Method [U-Solve]:
[tex]\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \leftarrow \\\end{aligned}[/tex]
[Integrand] Simplify:
[tex]\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \leftarrow \\\end{aligned}[/tex]
[Integral] Apply Arctrigonemtric Integration:
[tex]\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \leftarrow \\\end{aligned}[/tex]
Simplify:
[tex]\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \leftarrow \\\end{aligned}[/tex]
[u] Back-substitute:
[tex]\displaystyle \begin{aligned}\int {\frac{\sec x \tan x}{1 + \sec^2 x}} \, dx & = \int {\frac{\sec x \tan x \cos x \cot x}{u^2 + 1}} \, du \\& = \int {\frac{1}{u^2 + 1}} \, du \\& = \frac{1}{1} \arctan \bigg( \frac{u}{1} \bigg) + C \\& = \arctan u + C \\& = \boxed{ \arctan \big( \sec x \big) + C } \\\end{aligned}[/tex]

∴ we used substitution to find the indefinite integral.

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Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration