Question

\( \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx = \)

• A. $ 0 $
• B. $ \frac{2}{15} $
• C. $ \frac{4}{15} $
• D. $ \frac{2}{5} $

Hint:

When evaluating definite integrals over symmetric intervals, check if the integrand is odd or even. Odd functions integrated over symmetric intervals yield zero, simplifying calculations. Use substitutions like $ u = \cos x $ or $ v = \sin x $ to simplify powers of trigonometric functions.

Answer 1

The Correct Option is C

Step 1: Analyze the integral.
We need to evaluate: $$ I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx. $$ Split the integral into two parts: $$ I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x \, dx + \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \cos x \, dx. $$ Let: $$ I_1 = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x \, dx, \quad I_2 = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \cos x \, dx. $$ Thus: $$ I = I_1 + I_2. $$ Step 2: Evaluate $ I_1 $.
For $ I_1 $: $$ I_1 = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x \, dx. $$ Use the substitution $ u = \cos x $. Then: $$ du = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -du. $$ When $ x = -\frac{\pi}{2} $, $ u = \cos\left(-\frac{\pi}{2}\right) = 0 $. When $ x = \frac{\pi}{2} $, $ u = \cos\left(\frac{\pi}{2}\right) = 0 $. Thus, the limits of integration become $ u = 0 $ to $ u = 0 $. However, observe that $ \sin^2 x \cos^2 x \sin x $ is an odd function because: $$ \sin^2(-x) \cos^2(-x) \sin(-x) = (-\sin x)^2 (\cos x)^2 (-\sin x) = -\sin^2 x \cos^2 x \sin x. $$ Since the integrand is odd and the limits of integration are symmetric about zero, the integral evaluates to zero: $$ I_1 = 0. $$ Step 3: Evaluate $ I_2 $.
For $ I_2 $: $$ I_2 = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \cos x \, dx. $$ Use the substitution $ v = \sin x $. Then: $$ dv = \cos x \, dx \quad \Rightarrow \quad \cos x \, dx = dv. $$ When $ x = -\frac{\pi}{2} $, $ v = \sin\left(-\frac{\pi}{2}\right) = -1 $. When $ x = \frac{\pi}{2} $, $ v = \sin\left(\frac{\pi}{2}\right) = 1 $. Thus, the limits of integration become $ v = -1 $ to $ v = 1 $. The integral becomes: $$ I_2 = \int_{-1}^{1} (1 - v^2)v^2 \, dv = \int_{-1}^{1} (v^2 - v^4) \, dv. $$ Split the integral: $$ I_2 = \int_{-1}^{1} v^2 \, dv - \int_{-1}^{1} v^4 \, dv. $$ Evaluate each term separately: 1. For $ \int_{-1}^{1} v^2 \, dv $: $$ \int_{-1}^{1} v^2 \, dv = \left[ \frac{v^3}{3} \right]_{-1}^{1} = \frac{1^3}{3} - \frac{(-1)^3}{3} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3}. $$ 2. For $ \int_{-1}^{1} v^4 \, dv $: $$ \int_{-1}^{1} v^4 \, dv = \left[ \frac{v^5}{5} \right]_{-1}^{1} = \frac{1^5}{5} - \frac{(-1)^5}{5} = \frac{1}{5} + \frac{1}{5} = \frac{2}{5}. $$ Thus: $$ I_2 = \frac{2}{3} - \frac{2}{5} = \frac{10}{15} - \frac{6}{15} = \frac{4}{15}. $$ Step 4: Combine results.
Since $ I_1 = 0 $ and $ I_2 = \frac{4}{15} $, we have: $$ I = I_1 + I_2 = 0 + \frac{4}{15} = \frac{4}{15}. $$ Step 5: Final Answer.
$$ \boxed{\frac{4}{15}} $$