Concept Explanation: Evaluating a Definite Integral with Algebraic Manipulation

This integral appears complex due to the radical expression in the denominator. The key is to simplify the integrand by rewriting it in a form that allows for substitution. Notice that the denominator is a fourth root of a product of terms. We can express it as a power and then simplify.

Step 1: Rewrite the Denominator

The denominator is: ${\left({(x+1)}^2{(1-x)}^6\right)}^{1/4}$.

Using the property of exponents, ${(a\cdot b)}^n=a^n\cdot b^n$, we can write this as:

${(x+1)}^{2/4}\cdot {(1-x)}^{6/4}={(x+1)}^{1/2}\cdot {(1-x)}^{3/2}$.

Thus, the integrand becomes: $\frac{1+\sqrt{3}}{{(x+1)}^{1/2}\cdot {(1-x)}^{3/2}}$.

Step 2: Factor and Simplify the Expression

We can combine the terms in the denominator:

$\frac{1+\sqrt{3}}{{\left(\frac{1-x}{1+x}\right)}^{3/2}}$. However, a more straightforward substitution is often used for integrals containing $\sqrt{\frac{1-x}{1+x}}$.

Step 3: Apply a Trigonometric Substitution (or an Inverse Substitution)

Let $x=\cos 2\theta$. This is a common trick for such integrals because it simplifies expressions involving $\sqrt{1-x}$ and $\sqrt{1+x}$ using trigonometric identities.

Then: $1-x=1-\cos 2\theta =2{\sin }^2\theta$
$1+x=1+\cos 2\theta =2{\cos }^2\theta$
$\mathrm{dx}=-2\sin 2\theta d\theta$

Also, the limits change. When $x=0$, $\cos 2\theta =0\to \theta =\pi /4$. When $x=1/2$, $\cos 2\theta =1/2\to \theta =\pi /6$.

Step 4: Substitute into the Integral

The original integral $I$ becomes: $I=(1+\sqrt{3})\underset{\theta =\pi /4}{\overset{\theta =\pi /6}{\int }}\frac{-2\sin 2\theta d\theta }{\sqrt{2{\cos }^2\theta }\cdot {\left(2{\sin }^2\theta \right)}^{3/2}}$

Simplify the denominator: $\sqrt{2}|\cos \theta |\cdot {\left(2\right)}^{3/2}\left|{\sin }^3\theta \right|=\sqrt{2}\cdot 2\sqrt{2}|\cos \theta {\sin }^3\theta |=4|\cos \theta {\sin }^3\theta |$. Since we are integrating between 0 and π/2, cosθ and sinθ are positive, so we can drop the absolute values.

Also,