Question

If \(\int _{-π/2}^{π/2}(sin^2x+sin^3x)dx = k\), then the value of \(k\)

• A. \(0\)
• B. \(1\)
• C. \(π\)
• D. \(\frac{π}{2}\)

Hint:

The integral of an odd function over a symmetric interval is zero.

Answer 1

The Correct Option is D

Step 1: Approach
Use the Wallis-type integral for sine squared and symmetry.

Step 2: Odd term
$\sin^3x$ is odd, so contributes $0$.

Step 3: Even term
Over $[0,\pi/2]$, $\int\sin^2x\,dx=\dfrac{\pi}{4}$. Doubling for the full symmetric interval gives $\dfrac\pi2$.

Step 4: Result
$k=\dfrac\pi2$, option (D).

Final Answer:
The sin cubed term vanishes by symmetry and the sin squared term gives pi/2, option (D). \[ \boxed{\frac{\pi}{2}} \]