Question

The value of $\int_{-3}^{3} \sin^7 x \cos^{16} x \, dx$ is

• A. 1
• B. 2
• C. 0
• D. -1

Hint:

Check if the function is odd when you see limits like $\int_{-a}^{a}$; it often saves tedious calculation!

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The integral has symmetric limits from $-a$ to $a$. In such cases, checking whether the integrand is an even or odd function is the most efficient approach.
Step 2: Key Formula or Approach:
If $f(x)$ is an odd function (i.e., $f(-x) = -f(x)$), then $\int_{-a}^{a} f(x) dx = 0$.
If $f(x)$ is an even function (i.e., $f(-x) = f(x)$), then $\int_{-a}^{a} f(x) dx = 2\int_{0}^{a} f(x) dx$.
Step 3: Detailed Explanation:
Let the integrand be $f(x) = \sin^7 x \cos^{16} x$.
Replace $x$ with $-x$ to check the parity of the function:
\[ f(-x) = \sin^7(-x) \cos^{16}(-x) \] We know that sine is an odd function ($\sin(-\theta) = -\sin \theta$) and cosine is an even function ($\cos(-\theta) = \cos \theta$).
\[ f(-x) = (-\sin x)^7 (\cos x)^{16} \] Since the power of sine is 7 (an odd integer), the negative sign remains:
\[ f(-x) = -(\sin^7 x) (\cos^{16} x) \] \[ f(-x) = -f(x) \] Because $f(-x) = -f(x)$, the integrand is an odd function.
Therefore, by the property of definite integrals over symmetric limits, the integral evaluates to zero.
Step 4: Final Answer:
The value of the integral is 0.