Q.$\int_{-\pi/4}^{\pi/4} \log(\sin x+\cos x)\,dx$

The key idea is to use the property of definite integrals with symmetric limits:

$\int_{-a}^{a} f(x)\,dx = \int_{0}^{a} [f(x) + f(-x)]\,dx$.

Step 1: Let $I = \int_{-\pi/4}^{\pi/4} \log(\sin x + \cos x)\,dx$.

Replace $x$ by $-x$ in the integrand:

$f(-x) = \log(\sin(-x) + \cos(-x)) = \log(-\sin x + \cos x)$.

Step 2: Add $f(x)$ and $f(-x)$:

$f(x) + f(-x) = \log(\sin x + \cos x) + \log(\cos x - \sin x)$

$= \log[(\cos x + \sin x)(\cos x - \sin x)] = \log(\cos^2 x - \sin^2 x) = \log(\cos 2x)$.

Step 3: Using the symmetry property:

$I = \int_{0}^{\pi/4} \log(\cos 2x)\,dx$.

Substitute $t = 2x$, so $dx = dt/2$, limits: $0$ to $\pi/2$: …

The key insight here is that the limits are symmetric about zero, from $-\pi/4$ to $\pi/4$. When you see symmetric limits, your first instinct should be to check if the integrand has any symmetry — even, odd, or something that simplifies when you replace $x$ with $-x$. Here, the integrand is $\log(\sin x + \cos x)$, which is neither even nor odd. But the symmetry trick still works: we can rewrite the integral as half the sum of the function and its reflection.

Let’s walk through it.

1. Apply the symmetry property for definite integrals. For any function $f(x)$ integrated over $[-a, a]$, we have:

This is because the integral from $-a$ to $0$ can be transformed by substituting $x \to -x$, and then adding it to the integral from $0$ to $a$.

Here, $a = \pi/4$ and $f(x) = \log(\sin x + \cos x)$. So:

2. Simplify $f(-x)$. Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$, we get:

So the sum inside the integral becomes:

3. Use the identity $(\sin x + \cos x)(\cos x - \sin x) = \cos^2 x - \sin^2 x = \cos 2x$. This is a standard double-angle identity. So:

4. Substitute to simplify further. Let $t = 2x$. Then $dx = dt/2$, and when $x = 0$, $t = 0$; when $x = \pi/4$, $t = \pi/2$. So:

5. Recall the standard result for $\int_{0}^{\pi/2} \log(\cos t) \, dt$. This is a well-known integral. One way to derive it is to use the identity $\int_{0}^{\pi/2} \log(\sin t) \, dt = \int_{0}^{\pi/2} \log(\cos t) \, dt$ (by substituting $t \to \pi/2 - t$), and then note that:

But the left side, with $u = 2t$, becomes $\frac{1}{2} \int_{0}^{\pi} \log(\sin u) \, du$, and using symmetry, that equals $\int_{0}^{\pi/2} \log(\sin t) \, dt$. Solving gives:

Method: Symmetric limits via $f(x)+f(-x)$

When the limits are $[-a,a]$ but the integrand is neither plainly even nor odd, use $\int_{-a}^a f(x)\,dx=\int_0^a[f(x)+f(-x)]\,dx$.

Steps

Step 1: Form $f(x)+f(-x)$.

Replace $x$ by $-x$ using $\sin(-x)=-\sin x$, $\cos(-x)=\cos x$, and add. For log integrands, $\log P+\log Q=\log(PQ)$ often collapses to a simple product.

Step 2: Simplify the combined integrand. …

Common Mistakes

Mistake 1: Assuming $\log(\sin x+\cos x)$ is even or odd.

Why it's wrong: it is neither, so neither the "double it" nor the "it's zero" shortcut applies. Correct approach: use $\int_{-a}^a f=\int_0^a[f(x)+f(-x)]\,dx$, which turns the sum into $\log\cos 2x$.

Mistake 2: Dropping the $\frac12$ from the substitution $t=2x$. …