Question

Evaluate \[ \int_{0}^{\sqrt{\pi}} x\sin^{2}(x^{2})\,dx \]

• A. \(\dfrac{\pi}{4}\)
• B. \(\pi\)
• C. \(\dfrac{\sqrt{\pi}}{4}\)
• D. \(\sqrt{\pi}\)

Hint:

For integrals involving \(x\,f(x^2)\), immediately try the substitution \(t=x^2\). It often converts the problem into a standard trigonometric or exponential integral.

Answer 1

The Correct Option is A

Concept: Whenever an integrand contains \(x\) together with a function of \(x^2\), substitution \[ t=x^2 \] is usually the most effective technique.

Step 1: Perform substitution. Let \[ t=x^2 \] Then \[ dt=2x\,dx \] or \[ x\,dx=\frac{dt}{2} \] Limits change as: \[ x=0 \Rightarrow t=0 \] \[ x=\sqrt{\pi}\Rightarrow t=\pi \] Hence, \[ I = \int_{0}^{\sqrt{\pi}} x\sin^{2}(x^{2})\,dx \] becomes \[ I = \frac12 \int_{0}^{\pi} \sin^{2}t\,dt \]

Step 2: Use the standard identity. \[ \sin^{2}t = \frac{1-\cos2t}{2} \] Therefore, \[ I = \frac12 \int_{0}^{\pi} \frac{1-\cos2t}{2} \,dt \] \[ = \frac14 \int_{0}^{\pi} (1-\cos2t)\,dt \]

Step 3: Integrate. \[ I = \frac14 \left[ t-\frac{\sin2t}{2} \right]_{0}^{\pi} \] Substituting limits: \[ I = \frac14 \left[ \pi-\frac{\sin2\pi}{2} \right] \] \[ = \frac14(\pi) \] \[ I=\frac{\pi}{4} \]

Final Answer: \[ \boxed{\frac{\pi}{4}} \] Hence option (A) is correct.