Question
Find: $\int \cos ^{2} x d x$
✓
Answer
$\cos 2x = 2 \cos^2 x - 1$
$\cos ^{2} x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^{2} x d x=\frac{1}{2} \int(1+\cos 2 x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$
$=\frac{x}{2}+\frac{1}{4} \sin 2 x+C$
$\cos ^{2} x=\frac{1+\cos 2 x}{2}$
Therefore, $\int \cos ^{2} x d x=\frac{1}{2} \int(1+\cos 2 x) d x=\frac{1}{2} \int d x+\frac{1}{2} \int \cos 2 x d x$
$=\frac{x}{2}+\frac{1}{4} \sin 2 x+C$
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