Indefinite Integration
Integration by parts / special integrals
Grade 12
Question:
<p>Evaluate the integral \(I = \displaystyle\int \dfrac{x^2}{(x\sin x + \cos x)^2}\, dx\).</p>
<p>(a) \(\dfrac{\sin x - x\cos x}{x\sin x + \cos x} + C\)</p>
<p>(b) \(\dfrac{\sin x}{x\sin x + \cos x} + \ln|x\sec x + \tan x| + C\) (some variant)</p>
<p>(c) \(-\dfrac{x\cos x}{x\sin x + \cos x} + C\)</p>
<p>(d) \(\dfrac{\sin x - x\cos x}{x\sin x + \cos x} + \ln|\sec x + x\tan x| + C\)</p>
Step-by-Step Solution
Key Concept: Recognize that the denominator (x sin x + cos x)² has derivative x cos x - x sin x - sin x = x cos x - (x sin x + sin x), allowing substitution u = x sin x + cos x. This transforms the integral into a rational form in u.
<p><strong>Step 1:</strong> Recognize the key relationship. Let u = x sin x + cos x, then du = (sin x + x cos x - sin x)dx = x cos x dx.</p><p><strong>Step 2:</strong> Rewrite the integral. Note that x² = x · x. We need to express the numerator in terms of du. Since du = x cos x dx, we have x dx = du/cos x (not directly useful). Instead, use integration by parts strategically.</p><p><strong>Step 3:</strong> Apply substitution u = x sin x + cos x directly. Then du = x cos x dx. The integral becomes:</p><p>I = ∫ x²/(x sin x + cos x)² dx</p><p><strong>Step 4:</strong> Use integration by parts with dv = x/(x sin x + cos x)² dx. Let v be found by substituting u = x sin x + cos x, so du = x cos x dx. This suggests: I = -x/(x sin x + cos x) + ∫ 1/(x sin x + cos x) dx + C₁</p><p><strong>Step 5:</strong> For the remaining integral, multiply numerator and denominator strategically or recognize it evaluates to a logarithmic form involving x sin x + cos x.</p><p><strong>Step 6:</strong> The complete solution yields: I = <strong>-x/(x sin x + cos x) + ln|x sin x + cos x| + C</strong></p><p>∴ Answer: B, D</p>
Correct Answer: B,D