Animated Solution for Mathematics - Definite Integration: The value of $I={\int }_0^{\pi /2}\frac{(\sin x+\cos x{)}^2}{\sqrt{1+\sin 2x}}dx$ is

We are tasked with evaluating the integral:

When you see a complex trigonometric expression inside an integral, your first instinct should be to look for hidden symmetries and identities.

The denominator, $\sqrt{1+\sin 2x}$, is the key to unlocking this problem. We utilize the fundamental Pythagorean identity $1={\sin }^{2}x+{\cos }^{2}x$ and the double-angle formula $\sin 2x=2\sin x\cos x$.

Substituting these into the expression, we obtain:

This is the classic algebraic expansion $(a+b{)}^2=a^2+b^2+2ab$. Thus, we find:

We must be careful when taking the square root, as $\sqrt{u^2}=|u|$. Many students incorrectly write $\sqrt{(\sin x+\cos x{)}^2}=\sin x+\cos x$ without verification.

However, we are integrating over the interval $x\in [0,\pi /2]$. In this first quadrant, both $\sin x$ and $\cos x$ are non-negative, meaning their sum is strictly non-negative.

Therefore, we can safely drop the modulus bars:

With the denominator simplified to $\sin x+\cos x$, our integral becomes:

The terms cancel out, leaving us with a straightforward integral:

The antiderivative of $\sin x$ is $-\cos x$, and the antiderivative of $\cos x$ is $\sin x$. Thus, the primitive function is $F(x)=-\cos x+\sin x$.

Applying the Fundamental Theorem of Calculus, we evaluate $F(x)$ at the limits $0$ and $\pi /2$:

The final value of the integral is: