Question
Evaluate the following integrals:
$\int\text{x}\cos^2\text{x dx}$
$\int\text{x}\cos^2\text{x dx}$
✓
Answer
Let $\text{I}=\int\text{x}\cos^2\text{x dx}$
Using integration by parts,
$\text{I}=\text{x}\int\cos^2\text{x dx}-\int(1\int\cos^2\text{x dx})\text{dx}$
$=\text{x}\int\Big(\frac{\cos2\text{x}+1}{2}\Big)\text{dx}-\int\bigg(\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\bigg)\text{dx}$
$=\frac{\text{x}}{2}\Big[\frac{\sin2\text{x}}{2}+\text{x}\Big]-\frac{1}{2}\int\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{1}{4}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{4}+\frac{1}{8}\cos2\text{x}+\text{C}$
Using integration by parts,
$\text{I}=\text{x}\int\cos^2\text{x dx}-\int(1\int\cos^2\text{x dx})\text{dx}$
$=\text{x}\int\Big(\frac{\cos2\text{x}+1}{2}\Big)\text{dx}-\int\bigg(\int\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\bigg)\text{dx}$
$=\frac{\text{x}}{2}\Big[\frac{\sin2\text{x}}{2}+\text{x}\Big]-\frac{1}{2}\int\Big(\text{x}+\frac{\sin2\text{x}}{2}\Big)\text{dx}$
$=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{2}-\frac{1}{2}\times\frac{\text{x}^2}{2}-\frac{1}{4}\Big(-\frac{\cos2\text{x}}{2}\Big)+\text{C}$
$\text{I}=\frac{\text{x}}{4}\sin2\text{x}+\frac{\text{x}^2}{4}+\frac{1}{8}\cos2\text{x}+\text{C}$
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