Animated Solution for Mathematics - Definite Integration: The value of the integral ${\int }_{-1}^1{\log }_e(\sqrt{1-x}+\sqrt{1+x})dx$ is equal to :

A Journey Through Calculus

Welcome, fellow traveler on the road to JEE Advanced excellence. Today, we are going to dissect a problem that, at first glance, might seem like a daunting thicket of square roots and logarithms.

But as we peel back the layers, you will see that this problem is a masterclass in elegance and strategic thinking. Let us begin with the integral:

When you see limits of the form $[-a,a]$, your mathematical spider-sense should immediately tingle. This is the universe whispering to you: "Check for symmetry."

The Mirror Image

Let $f(x)={\log }_e(\sqrt{1-x}+\sqrt{1+x})$. To check for symmetry, we perform the substitution $x\to -x$.

As we calculate $f(-x)={\log }_e(\sqrt{1-(-x)}+\sqrt{1+(-x)})$, we see the terms inside the square roots simply swap places:

Because addition is commutative, $f(-x)=f(x)$. Our function is perfectly even, meaning its graph is a mirror image across the $y$-axis.

This allows us to invoke the property ${\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx$. Our integral now becomes:

We have successfully cut our work in half and anchored our lower limit at zero. This is the first victory.

The Tool of Integration by Parts

Now, we face the challenge of integrating a lone logarithmic function. The standard tool in our arsenal is Integration by Parts (IBP), guided by the ILATE rule.

We set $u={\log }_e(\sqrt{1-x}+\sqrt{1+x})$ and $dv=dx$. Consequently, $v=x$.

The real test lies in finding $du$. We must use the chain rule:

This yields:

I know this looks terrifying, but take a breath. We are about to perform an algebraic dance to simplify this.

The Algebraic Masterstroke

To simplify $du$, we take the lowest common multiple inside the bracket:

Now, we rationalize the expression by multiplying the numerator and denominator by the conjugate $(\sqrt{1-x}-\sqrt{1+x})$. After expanding and canceling, we arrive at the much neater:

This simplification is the heart of the problem. When we plug this into the IBP formula $I=2[uv{]}_0^1-2{\int }_0^{1}vdu$, the $x$ in our $v$ term and the $x$ in the denominator of $du$ cancel out perfectly!

We are left with:

The Final Integration

Evaluating the boundary term is straightforward. At $x=1$, we get $2{\log }_e(\sqrt{2})={\log }_{e}2$. At $x=0$, the term vanishes.

Now, for the remaining integral:

These are standard, elementary integrals. The first part is ${\sin }^{-1}x$, and the second is $x$.

Evaluating from $0$ to $1$, we get $({\sin }^{-1}(1)-1)-(0-0)=\frac{\pi }{2}-1$.

Combining our boundary term and our integral, we reach the final answer:

You have navigated the symmetry, mastered the IBP, and conquered the algebra. This is the essence of JEE Advanced mathematics—not just calculation, but the art of simplification.