Question

The value of \(\displaystyle\int_{0}^{\pi/4}\sin^{3}2x\cos2x\,dx\) is:

• A. \(\dfrac12\)
• B. \(\dfrac14\)
• C. \(\dfrac18\)
• D. \(\dfrac1{16}\)

Hint:

Substitute u = sin2x.

Answer 1

The Correct Option is C

Step 1: Key Substitution:
Let \(u=\sin2x\), so \(du=2\cos2x\,dx\), i.e. \(\cos2x\,dx=\dfrac{du}{2}\).

Step 2: Rewriting the integral:
\(\displaystyle\int\sin^{3}2x\cos2x\,dx=\int u^{3}\cdot\frac{du}{2}=\frac{u^{4}}{8}=\frac{\sin^{4}2x}{8}\).

Step 3: Applying the limits:
At \(x=\pi/4\): \(\sin(\pi/2)=1\), giving \(\dfrac{1}{8}\). At \(x=0\): \(\sin0=0\), giving \(0\).

Final Answer:
\[ \boxed{\dfrac18} \]