Question

The value of the integral \( \int_{-\pi}^{\pi} (\cos^6 x + \cos^4 x) \, dx \) is:

• A. $\frac{\pi}{2}$
• B. $\frac{5\pi}{8}$
• C. $\frac{11\pi}{8}$
• D. $\frac{9\pi}{8}$

Hint:

Remember the properties of even functions when integrating over symmetric intervals. The reduction formula for $\int_{0}^{\pi/2} \cos^n x dx$ is essential for solving integrals of powers of cosine.

Answer 1

The Correct Option is C

Let the given integral be $I$.
$$I = \int_{-\pi}^{\pi} (\cos^6 x + \cos^4 x) dx$$ Since $\cos x$ is an even function, $\cos^6 x$ and $\cos^4 x$ are also even functions. Therefore,
$$I = 2 \int_{0}^{\pi} (\cos^6 x + \cos^4 x) dx = 2 \left( \int_{0}^{\pi} \cos^6 x dx + \int_{0}^{\pi} \cos^4 x dx \right)$$ We use the property $\int_{0}^{\pi} \cos^n x dx = 2 \int_{0}^{\pi/2} \cos^n x dx$ for even $n$. For $n = 6$:
$$\int_{0}^{\pi} \cos^6 x dx = 2 \int_{0}^{\pi/2} \cos^6 x dx$$ Using the reduction formula $\int_{0}^{\pi/2} \cos^n x dx = \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{1}{2} \frac{\pi}{2}$ for even $n$: $$\int_{0}^{\pi/2} \cos^6 x dx = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{15\pi}{96} = \frac{5\pi}{32}$$ So, $\int_{0}^{\pi} \cos^6 x dx = 2 \cdot \frac{5\pi}{32} = \frac{5\pi}{16}$. For $n = 4$:
$$\int_{0}^{\pi} \cos^4 x dx = 2 \int_{0}^{\pi/2} \cos^4 x dx$$ Using the reduction formula:
$$\int_{0}^{\pi/2} \cos^4 x dx = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$$ So, $\int_{0}^{\pi} \cos^4 x dx = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$. Now, substitute these values back into the expression for $I$:
$$I = 2 \left( \frac{5\pi}{16} + \frac{3\pi}{8} \right) = 2 \left( \frac{5\pi}{16} + \frac{6\pi}{16} \right) = 2 \left( \frac{11\pi}{16} \right) = \frac{11\pi}{8}$$