Question

\( \int_{0}^{\frac{\sqrt{\pi}}{2}} 2x^3 \sin(x^2)\, dx = \)

• A. \( \frac{1}{\sqrt{2}}\left(1+\frac{\pi}{4}\right) \)
• B. \( \frac{1}{\sqrt{2}}\left(1-\frac{\pi}{4}\right) \)
• C. \( \frac{1}{\sqrt{2}}\left(\frac{\pi}{2}-1\right) \)
• D. \( \frac{1}{\sqrt{2}}\left(1-\frac{\pi}{2}\right) \)
• E. \( \frac{1}{\sqrt{2}}\left(\frac{\pi}{4}-1\right) \)

Hint:

Whenever you see \(x^n \sin(x^2)\), substitution \(t=x^2\) simplifies the integral immediately.

Answer 1

The Correct Option is B

Concept: Use substitution for composite functions: \[ t = x^2 \Rightarrow dt = 2x\,dx \]

Step 1: Substitute
Let: \[ t = x^2,\quad dt = 2x dx \] Then: \[ 2x^3 dx = x^2 \cdot 2x dx = t\, dt \]

Step 2: Change limits
\[ x=0 \Rightarrow t=0,\quad x=\frac{\sqrt{\pi}}{2} \Rightarrow t=\frac{\pi}{4} \]

Step 3: Transform integral
\[ \int_0^{\pi/4} t \sin t \, dt \]

Step 4: Integration by parts
Let: \[ u = t,\quad dv = \sin t dt \] \[ du = dt,\quad v = -\cos t \] \[ \int t\sin t dt = -t\cos t + \int \cos t dt \] \[ = -t\cos t + \sin t \]

Step 5: Apply limits
\[ \left[-t\cos t + \sin t\right]_0^{\pi/4} \] \[ = -\frac{\pi}{4}\cos\frac{\pi}{4} + \sin\frac{\pi}{4} \] \[ = -\frac{\pi}{4}\cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \] \[ = \frac{\sqrt{2}}{2}\left(1 - \frac{\pi}{4}\right) \]

Step 6: Final Answer
\[ \boxed{\frac{1}{\sqrt{2}}\left(1 - \frac{\pi}{4}\right)} \]