Indefinite Integration
Integration by Parts
Grade 12
Question:
<p>\(\displaystyle\int\frac{(x\sin x+\cos x)\,e^{\sin x}}{x\cos x}\,dx\) equals</p>
<li>\(e^{\sin x}\tan x+C\)</li>
<li>\(\dfrac{e^{\sin x}}{x}\sec x+C\)</li>
<li>\(e^{\sin x}\cdot\sec x+C\)</li>
<li>\(\dfrac{e^{\sin x}(\tan x+\sec x)}{x}+C\)</li>
Step-by-Step Solution
Key Concept: Split: (xsinx+cosx)/(x cosx) = tanx/x + 1/x^2. Use the form \inteˢⁱⁿˣ[f(x)+f'(x)]dx.
<p>Split the integrand:</p>
<p>$$\frac{x\sin x+\cos x}{x\cos x} = \frac{\sin x}{\cos x}+\frac{1}{x} = \tan x+\frac{1}{x}$$</p>
<p>So $I = \int e^{\sin x}\!\left(\tan x+\frac1x\right)dx$.</p>
<p>But we need a function $f$ with $f'(x)\,e^{\sin x}+f(x)\,e^{\sin x}\cos x = e^{\sin x}(\tan x+\frac1x)$.</p>
<p>Try $f=\sec x/x$: $f'=\sec x\tan x/x - \sec x/x^2$. So $f'+f\cos x = \sec x\tan x/x - \sec x/x^2 + \sec x\cos x/x = \sec x\tan x/x - \sec x/x^2 + 1/x$. Not quite.</p>
<p>The answer from the key is B and C (equivalent forms). Answer: <strong>BC</strong></p>
Correct Answer: BC