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: Understanding the Concept
For an interval symmetric about zero, an odd function integrates to zero and an even function integrates to twice the integral over half the interval.

Step 2: Split
\[ k=\int_{-\pi/2}^{\pi/2}\sin^2x\,dx+\int_{-\pi/2}^{\pi/2}\sin^3x\,dx \]

Step 3: The sin cubed part
\(\sin^3(-x)=-\sin^3x\), so it is odd and its integral is 0.

Step 4: The sin squared part
\(\sin^2x\) is even:
\[ \int_{-\pi/2}^{\pi/2}\sin^2x\,dx=2\int_0^{\pi/2}\frac{1-\cos2x}{2}dx=\left[x-\frac{\sin2x}{2}\right]_0^{\pi/2}=\frac\pi2 \]

Step 5: Total
\[ k=\frac\pi2+0=\frac\pi2 \]
This is option (D).

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