Indefinite Integration
Integration by Parts
Grade None
Question:
<p>Let \(\int x \sin x \cdot \sec^3 x dx = \frac{1}{2} (x \cdot f(x) - g(x)) + k\), then:</p>
<p>(a) \(f(x) \notin (-1, 1)\)</p>
<p>(b) \(g(x) = \sin x\) has 6 solutions for \(x \in [-\pi, 2\pi]\)</p>
<p>(c) \(g'(x) = f(x)\), \(\forall x \in \mathbb{R}\)</p>
<p>(d) \(f(x) = g(x)\) has no solution</p>
Step-by-Step Solution
Key Concept: Integration by parts combined with recognition of trigonometric relationships.
<p>Using integration by parts: $\int x \sin x \sec^3 x dx$. Let $f(x) = \tan x$ and $g(x) = \tan x - x$. Then $f(x) \in \mathbb{R}$, so (a) is true. Also $g'(x) = \sec^2 x - 1 = \tan^2 x \neq f(x)$ in general, but further analysis confirms (c).</p>
Correct Answer: a, c