8. Find, in simplest form,
$$
\int(2 \cos x-\sin x)^{2} d x
$$

Question

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ We start by expanding the integrand $(2 \cos x - \sin x)^2$. ‖ ‖ step 2 ⋮ Expanding $(2 \cos x - \sin x)^2$ gives $4 \cos^2 x - 4 \cos x \sin x + \sin^2 x$. ‖ ‖ step 3 ⋮ We use trigonometric identities to simplify: $\cos^2 x = \frac{1 + \cos 2x}{2}$ and $\sin^2 x = \frac{1 - \cos 2x}{2}$. ‖ ‖ step 4 ⋮ Substituting these identities, we get $4 \left(\frac{1 + \cos 2x}{2}\right) - 4 \cos x \sin x + \left(\frac{1 - \cos 2x}{2}\right)$. ‖ ‖ step 5 ⋮ Simplifying further, we obtain $2 (1 + \cos 2x) - 4 \cos x \sin x + \frac{1 - \cos 2x}{2}$. ‖ ‖ step 6 ⋮ Combining like terms, we get $\frac{5}{2} + \frac{3}{2} \cos 2x - 4 \cos x \sin x$. ‖ ‖ step 7 ⋮ We now integrate each term separately: $\int \frac{5}{2} dx + \int \frac{3}{2} \cos 2x dx - \int 4 \cos x \sin x dx$. ‖ ‖ step 8 ⋮ Integrating each term, we get $\frac{5}{2} x + \frac{3}{4} \sin 2x + \cos 2x + C$. ‖ ∻Answer∻ ⚹ $\frac{5}{2} x + \frac{3}{4} \sin 2x + \cos 2x + C$ ⚹ ∻Key Concept∻ ⚹Trigonometric Identities and Integration⚹ ∻Explanation∻ ⚹We used trigonometric identities to simplify the integrand and then integrated each term separately.⚹◊What is the integral of $(2 \cos x - \sin x)^{2}$ with respect to $x$?⍭ Generate me a similar question◊

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