Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-\pi /2}^{\pi /2}(x^2+\ln \frac{\pi +x}{\pi -x})\cos xdx$ is

When you encounter an integral with symmetric limits, such as from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$, this is a massive, flashing neon sign in the world of JEE Advanced. Your first instinct should always be to hunt for symmetry.

Is the function even? Is it odd? Or is it a mix of both? This problem is a perfect case study in how to decompose a complex challenge into manageable, elegant pieces.

The Logarithmic Vanishing Act

Let us look at our integrand: $f(x)=(x^2+\ln \frac{\pi +x}{\pi -x})\cos x$. We can distribute the $\cos x$ to split this into two distinct integrals:

Now, focus on the second term. Let $g(x)=\ln \left(\frac{\pi +x}{\pi -x}\right)\cos x$. To test for symmetry, we replace $x$ with $-x$:

Since $\cos (-x)=\cos x$ and $\frac{\pi -x}{\pi +x}$ is the reciprocal of $\frac{\pi +x}{\pi -x}$, we can write this as:

We have proven that this entire logarithmic term is an odd function. Consequently, the integral of an odd function over symmetric limits vanishes. It is gone, leaving us with only the first part.

We are now left with $I={\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{x}^2\cos xdx$. Let $h(x)=x^2\cos x$. Testing for symmetry, $h(-x)=(-x{)}^2\cos (-x)=x^2\cos x=h(x)$.

This is an even function, which allows us to simplify the integral:

To solve this, we use Integration by Parts (IBP), where $\int udv=uv-\int vdu$. We choose $u=x^2$ and $dv=\cos xdx$, which gives $du=2xdx$ and $v=\sin x$.

Applying the formula:

We must apply IBP again to the remaining integral $\int 2x\sin xdx$. Choosing $u=2x$ and $dv=\sin xdx$, we get $du=2dx$ and $v=-\cos x$. The integral becomes:

Combining everything, our antiderivative is $x^2\sin x+2x\cos x-2\sin x$. We evaluate this from $0$ to $\frac{\pi }{2}$ and multiply by the $2$ we pulled out earlier:

At the upper limit $\frac{\pi }{2}$:

At the lower limit $0$, the expression evaluates to $0$. Multiplying by $2$, we arrive at the final result: