Question:
$\int_{-\pi/2}^{\pi/2}(x^{5}+x^{3}+x)\cos x dx = $ ________.
$\int_{-\pi/2}^{\pi/2}(x^{5}+x^{3}+x)\cos x dx = $ ________.
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$\int_{-a}^{a} (\text{odd function}) dx = 0$.
Updated On: Jun 26, 2026
- $\frac{\pi}{4}$
- $\pi$
- $\frac{2\pi}{3}$
- $\frac{\pi}{2}$
- 0
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The Correct Option is
Solution and Explanation
Step 1: Concept
Check if the integrand is an even or odd function over a symmetric interval $[-a, a]$.
Step 2: Meaning
An odd function $f(x)$ satisfies $f(-x) = -f(x)$. The integral of an odd function from $-a$ to $a$ is zero.
Step 3: Analysis
Let $f(x) = (x^5 + x^3 + x)\cos x$.
$f(-x) = ((-x)^5 + (-x)^3 + (-x))\cos(-x) = (-x^5 - x^3 - x)\cos x = -f(x)$.
Step 4: Conclusion
Since the function is odd, the integral is 0. Final Answer: (E)
Check if the integrand is an even or odd function over a symmetric interval $[-a, a]$.
Step 2: Meaning
An odd function $f(x)$ satisfies $f(-x) = -f(x)$. The integral of an odd function from $-a$ to $a$ is zero.
Step 3: Analysis
Let $f(x) = (x^5 + x^3 + x)\cos x$.
$f(-x) = ((-x)^5 + (-x)^3 + (-x))\cos(-x) = (-x^5 - x^3 - x)\cos x = -f(x)$.
Step 4: Conclusion
Since the function is odd, the integral is 0. Final Answer: (E)
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