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: Name the integral.
Let $I=\displaystyle\int_0^{\pi} x\,\big(\sin^2(\sin x)+\cos^2(\cos x)\big)\,dx$ and let $h(x)=\sin^2(\sin x)+\cos^2(\cos x)$.
Step 2: Use the reflection property.
Replace $x$ by $\pi-x$. Since $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$ with cosine squared, $h(\pi-x)=h(x)$, so $h$ is symmetric about $x=\dfrac{\pi}{2}$.
Step 3: Write the mirrored integral.
Then $I=\displaystyle\int_0^{\pi}(\pi-x)h(x)\,dx$.
Step 4: Add the two forms.
Adding, $2I=\displaystyle\int_0^{\pi}\big[x+(\pi-x)\big]h(x)\,dx=\pi\int_0^{\pi}h(x)\,dx$.
Step 5: Evaluate the symmetric integral.
Splitting and using the symmetry of $\sin^2(\sin x)$ and $\cos^2(\cos x)$ over $[0,\pi]$, their contributions combine so that $\displaystyle\int_0^{\pi}h(x)\,dx=\pi$. Hence $2I=\pi\cdot\pi=\pi^2$.
Step 6: Solve for $I$.
$I=\dfrac{\pi^2}{2}$.
\[ \boxed{\dfrac{\pi^2}{2}} \]