Question

\(\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} sin^2xcos^2x(sinx+cosx)dx=\)

• A. \(\frac{2}{3}\)
• B. \(\frac{3}{10}\)
• C. \(\frac{4}{15}\)
• D. \(\frac{5}{18}\)

Answer 1

The Correct Option is C

To solve the problem, we need to evaluate the integral $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x) dx$

1. Split the Integral:
Write $I = \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x dx + \int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \cos x dx$.

2. Analyze the First Integral:
Let $f(x) = \sin^2 x \cos^2 x \sin x$.
Check if $f(x)$ is odd:
$f(-x) = \sin^2(-x) \cos^2(-x) \sin(-x) = \sin^2 x \cos^2 x (-\sin x) = -f(x)$.
Since $f(x)$ is odd, $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x dx = 0$.

3. Analyze the Second Integral:
Let $g(x) = \sin^2 x \cos^2 x \cos x$.
Check if $g(x)$ is even:
$g(-x) = \sin^2(-x) \cos^2(-x) \cos(-x) = \sin^2 x \cos^2 x \cos x = g(x)$.
Since $g(x)$ is even, $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \cos x dx = 2 \int_0^{\pi/2} \sin^2 x \cos^2 x \cos x dx$.

4. Evaluate the Remaining Integral:
Thus, $I = 2 \int_0^{\pi/2} \sin^2 x \cos^2 x \cos x dx$.
Use $\cos^2 x = 1 - \sin^2 x$:
$I = 2 \int_0^{\pi/2} \sin^2 x (1 - \sin^2 x) \cos x dx$.

5. Substitute and Integrate:
Let $u = \sin x$, so $du = \cos x dx$. Limits: $x = 0 \to u = 0$, $x = \pi/2 \to u = 1$.
Then $I = 2 \int_0^1 u^2 (1 - u^2) du = 2 \int_0^1 (u^2 - u^4) du$.
Integrate:
$2 \left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1 = 2 \left( \frac{1}{3} - \frac{1}{5} \right) = 2 \cdot \frac{5 - 3}{15} = 2 \cdot \frac{2}{15} = \frac{4}{15}$.

Final Answer:
The value of the integral is $\frac{4}{15}$.