1. We are asked to find the integral of $\cos^6 x \, dx$. 2. Use the power-reduction formula for cosine: $$\cos^2 x = \frac{1 + \cos 2x}{2}$$ 3. Rewrite $\cos^6 x$ as $(\cos^2 x)^3$ and substitute: $$\cos^6 x = \left(\frac{1 + \cos 2x}{2}\right)^3$$ 4. Expand the cube: $$\left(\frac{1 + \cos 2x}{2}\right)^3 = \frac{1}{8} (1 + \cos 2x)^3 = \frac{1}{8} (1 + 3 \cos 2x + 3 \cos^2 2x + \cos^3 2x)$$ 5. Now the integral becomes: $$\int \cos^6 x \, dx = \frac{1}{8} \int (1 + 3 \cos 2x + 3 \cos^2 2x + \cos^3 2x) \, dx$$ 6. Integrate each term separately. - Integral of $1$ is $x$. - Integral of $3 \cos 2x$ is $\frac{3}{2} \sin 2x$ (since $\int \cos ax \, dx = \frac{\sin ax}{a}$). - For $3 \cos^2 2x$, use power reduction again: $$\cos^2 2x = \frac{1 + \cos 4x}{2}$$ So, $$3 \int \cos^2 2x \, dx = 3 \int \frac{1 + \cos 4x}{2} \, dx = \frac{3}{2} \int (1 + \cos 4x) \, dx = \frac{3}{2} \left(x + \frac{\sin 4x}{4}\right) = \frac{3}{2} x + \frac{3}{8} \sin 4x$$ - For $\int \cos^3 2x \, dx$, use the identity: $$\cos^3 \theta = \frac{3 \cos \theta + \cos 3\theta}{4}$$ Substitute $\theta = 2x$: $$\cos^3 2x = \frac{3 \cos 2x + \cos 6x}{4}$$ So, $$\int \cos^3 2x \, dx = \frac{1}{4} \int (3 \cos 2x + \cos 6x) \, dx = \frac{1}{4} \left(\frac{3}{2} \sin 2x + \frac{1}{6} \sin 6x \right) = \frac{3}{8} \sin 2x + \frac{1}{24} \sin 6x$$ 7. Combine all parts: $$\int \cos^6 x \, dx = \frac{1}{8} \left( x + \frac{3}{2} \sin 2x + \frac{3}{2} x + \frac{3}{8} \sin 4x + \frac{3}{8} \sin 2x + \frac{1}{24} \sin 6x \right) + C$$ 8. Simplify inside the parentheses: $$\left( (x + \frac{3}{2} x) + \left( \frac{3}{2} + \frac{3}{8} \right) \sin 2x + \frac{3}{8} \sin 4x + \frac{1}{24} \sin 6x \right) = \left( \frac{5}{2} x + \frac{15}{8} \sin 2x + \frac{3}{8} \sin 4x + \frac{1}{24} \sin 6x \right)$$ 9. Multiply everything by $\frac{1}{8}$: $$\frac{5}{16} x + \frac{15}{64} \sin 2x + \frac{3}{64} \sin 4x + \frac{1}{192} \sin 6x + C$$ **Final answer:** $$\int \cos^6 x \, dx = \frac{5x}{16} + \frac{15}{64} \sin 2x + \frac{3}{64} \sin 4x + \frac{1}{192} \sin 6x + C$$