Indefinite Integration
Integration by Parts
Grade 12
Question:
<p>\(\displaystyle\int \cos(100x)\cdot\sin^{95}x\,dx\) equals</p>
<li>\(\dfrac{\sin(100x)\cdot\sin^{95}x}{100}+\dfrac{95}{100}\displaystyle\int\sin(100x)\cdot\sin^{94}x\cos x\,dx+C\)</li>
<li>\(\dfrac{\sin(100x)\cdot\sin^{95}x}{100}-\dfrac{19}{20}\displaystyle\int\sin(100x)\cdot\sin^{94}x\cos x\,dx+C\)</li>
<li>\(\dfrac{\cos(100x)\cdot\sin^{96}x}{96}+C\)</li>
<li>\(\dfrac{\sin(100x)\cdot\sin^{96}x}{9600}+C\)</li>
Step-by-Step Solution
Key Concept: Integrate by parts: u=sin^9^5x, dv=cos(100x)dx. Then du=95sin^9^4x \cdot cosx dx, v=sin(100x)/100.
<p><strong>Integration by parts:</strong> $u=\sin^{95}x,\;dv=\cos(100x)\,dx$</p>
<p>$du = 95\sin^{94}x\cos x\,dx,\quad v=\dfrac{\sin(100x)}{100}$</p>
<p>$$\int\cos(100x)\sin^{95}x\,dx = \frac{\sin(100x)\sin^{95}x}{100} - \frac{95}{100}\int\sin(100x)\sin^{94}x\cos x\,dx$$</p>
<p>$$= \frac{\sin(100x)\sin^{95}x}{100} - \frac{19}{20}\int\sin(100x)\sin^{94}x\cos x\,dx+C$$</p>
<p>Answer: <strong>(B)</strong></p>
Correct Answer: B