Animated Solution for Mathematics - Definite Integration: Evaluate ${\int }_0^1\log [\sqrt{1-x}+\sqrt{1+x}]dx$

We are tasked with evaluating the integral:

At first glance, this expression appears complex. However, we can treat this as an application of Integration by Parts by setting the integrand as $1\cdot \log (\dots )$.

We define our variables as follows: $u=\log (\sqrt{1-x}+\sqrt{1+x})$ $dv=dx$

Applying the formula $\int udv=uv-\int vdu$, we first differentiate $u$ using the chain rule:

To simplify $du$, we combine the terms inside the bracket:

By multiplying the numerator and denominator by the conjugate $(\sqrt{1-x}-\sqrt{1+x})$, the expression simplifies significantly. The denominator becomes $(\sqrt{1-x}{)}^2-(\sqrt{1+x}{)}^2=(1-x)-(1+x)=-2x$.

After algebraic cancellation, the integral term $\int vdu$ transforms into:

The boundary term from Integration by Parts, $[x\log (\sqrt{1-x}+\sqrt{1+x}){]}_0^1$, evaluates to $\frac{1}{2}\log 2$.

Evaluating the remaining standard integrals: 1. $\int \frac{x}{2\sqrt{1-x^2}}dx=-\frac{1}{2}\sqrt{1-x^2}$ 2. $\int \frac{1}{2}dx=\frac{1}{2}x$

Combining these results over the interval $[0,1]$, we arrive at the final, elegant result:

This result is the reward for rigorous algebraic manipulation and systematic calculus. The final answer is $I=\frac{1}{2}\log 2+\frac{\pi }{4}-\frac{1}{2}$.