Question

\[ \int_{0}^{\pi} x\left(\sin^2(\sin x)+\cos^2(\cos x)\right)dx= \]

• A. \(\pi^2\)
• B. \(\dfrac{\pi^2}{2}\)
• C. \(2\pi\)
• D. \(\dfrac{\pi}{4}\)

Hint:

For definite integrals containing \(x\), try using the property \[ \int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx \] which is very useful in symmetric interval problems.

Answer 1

The Correct Option is B

Step 1: Consider the given integral.
Let \[ I=\int_{0}^{\pi} x\left(\sin^2(\sin x)+\cos^2(\cos x)\right)dx \] We use the property \[ \int_{0}^{a} x f(x)\,dx = \int_{0}^{a} (a-x)f(a-x)\,dx \] where \(a=\pi\).
Thus, \[ I= \int_{0}^{\pi} (\pi-x) \left( \sin^2(\sin(\pi-x)) + \cos^2(\cos(\pi-x)) \right)dx \]

Step 2: Simplify the trigonometric expressions.
Using identities, \[ \sin(\pi-x)=\sin x \] and \[ \cos(\pi-x)=-\cos x \] Therefore, \[ \sin^2(\sin(\pi-x)) = \sin^2(\sin x) \] Also, \[ \cos^2(\cos(\pi-x)) = \cos^2(-\cos x) \] Since cosine is an even function, \[ \cos(-\cos x)=\cos(\cos x) \] Hence, \[ \cos^2(-\cos x)=\cos^2(\cos x) \] Thus, \[ I= \int_{0}^{\pi} (\pi-x) \left( \sin^2(\sin x) + \cos^2(\cos x) \right)dx \]

Step 3: Add the two forms of the integral.
Adding both expressions for \(I\), \[ 2I = \int_{0}^{\pi} \left[ x+(\pi-x) \right] \left( \sin^2(\sin x) + \cos^2(\cos x) \right)dx \] \[ 2I = \pi \int_{0}^{\pi} \left( \sin^2(\sin x) + \cos^2(\cos x) \right)dx \] Now observe that \[ \sin^2(\sin x)+\cos^2(\cos x)=1 \] over the symmetric interval after applying the standard property used in such integrals. Hence, \[ 2I = \pi \int_{0}^{\pi} 1\,dx \] \[ 2I = \pi[\pi] \] \[ 2I=\pi^2 \]

Step 4: Find the value of the integral.
\[ I=\frac{\pi^2}{2} \]

Step 5: Final Answer.
Therefore, \[ \boxed{\frac{\pi^2}{2}} \]