Question

Let \(\alpha\) and \(\beta\), \((\alpha\lt \beta)\), are roots of \(18x^2-9\pi x+\pi^2=0\), \(f(x)=x^2\), \(g(x)=\cos x\). Then \(\displaystyle \int_{\alpha}^{\beta} x(gof(x))dx=\)

• A. \(\dfrac{\sqrt{3}-1}{4}\)
• B. \(\dfrac{\sqrt{3}}{4}\)
• C. \(\dfrac{2+\sqrt{3}}{2}\)
• D. \(\dfrac{1}{2}\left(\sin\dfrac{\pi^2}{9}-\sin\dfrac{\pi^2}{36}\right)\)

Hint:

For integrals of the form \(\int x\cos(x^2)\,dx\), use the substitution \(u=x^2\), so that \(du=2x\,dx\).

Answer 1

The Correct Option is D

Step 1: Understand the composite function.
Given, \[ f(x)=x^2 \] and \[ g(x)=\cos x \] Therefore, \[ (gof)(x)=g(f(x)) \] \[ =\cos(x^2) \] So, the given integral becomes \[ \int_{\alpha}^{\beta} x(gof(x))dx = \int_{\alpha}^{\beta} x\cos(x^2)\,dx \]

Step 2: Find the roots \(\alpha\) and \(\beta\).
The quadratic equation is \[ 18x^2-9\pi x+\pi^2=0 \] Using the quadratic formula, \[ x=\frac{9\pi\pm\sqrt{(-9\pi)^2-4(18)(\pi^2)}}{2(18)} \] \[ x=\frac{9\pi\pm\sqrt{81\pi^2-72\pi^2}}{36} \] \[ x=\frac{9\pi\pm\sqrt{9\pi^2}}{36} \] \[ x=\frac{9\pi\pm3\pi}{36} \] Thus, \[ x=\frac{12\pi}{36} \quad \text{or} \quad x=\frac{6\pi}{36} \] \[ x=\frac{\pi}{3} \quad \text{or} \quad x=\frac{\pi}{6} \] Since \(\alpha\lt \beta\), \[ \alpha=\frac{\pi}{6}, \quad \beta=\frac{\pi}{3} \]

Step 3: Evaluate the integral.
Now, \[ \int_{\alpha}^{\beta} x\cos(x^2)\,dx = \int_{\pi/6}^{\pi/3} x\cos(x^2)\,dx \] Let \[ u=x^2 \] Then, \[ du=2x\,dx \] So, \[ x\,dx=\frac{du}{2} \] Therefore, \[ \int x\cos(x^2)\,dx = \frac{1}{2}\int \cos u\,du \] \[ =\frac{1}{2}\sin u \] \[ =\frac{1}{2}\sin(x^2) \]

Step 4: Apply the limits.
\[ \int_{\pi/6}^{\pi/3} x\cos(x^2)\,dx = \left[\frac{1}{2}\sin(x^2)\right]_{\pi/6}^{\pi/3} \] \[ = \frac{1}{2} \left[ \sin\left(\frac{\pi}{3}\right)^2 - \sin\left(\frac{\pi}{6}\right)^2 \right] \] Since, \[ \left(\frac{\pi}{3}\right)^2=\frac{\pi^2}{9} \] and \[ \left(\frac{\pi}{6}\right)^2=\frac{\pi^2}{36} \] we get \[ = \frac{1}{2} \left( \sin\frac{\pi^2}{9} - \sin\frac{\pi^2}{36} \right) \]

Step 5: Final Answer.
Hence, \[ \boxed{ \frac{1}{2} \left( \sin\frac{\pi^2}{9} - \sin\frac{\pi^2}{36} \right) } \]