Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi}\log(1+\cos x)\,dx$

The key idea is to use the property of definite integrals $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$, combined with a known standard integral.

Step 1: Let $I = \int_{0}^{\pi} \log(1+\cos x) \, dx$.

Using the property with $a = \pi$:

$I = \int_{0}^{\pi} \log(1+\cos(\pi - x)) \, dx = \int_{0}^{\pi} \log(1 - \cos x) \, dx$.

Step 2: Add the two expressions for $I$: …

The trick here is that the integrand $\log(1+\cos x)$ doesn't have a nice antiderivative you can write down easily. But definite integrals have symmetry properties that let us transform the problem into something simpler. The key property is:

This works because as $x$ runs from $0$ to $a$, the quantity $a-x$ runs from $a$ to $0$, covering the same interval in reverse. The area under the curve doesn't care about direction.

Let's apply this to our integral.

1. Write the original integral. Let

2. Apply the symmetry property with $a = \pi$. Replace $x$ by $\pi - x$:

Now $\cos(\pi - x) = -\cos x$, so

3. Add the two expressions for $I$. We now have two different-looking integrals that are actually equal. Adding them gives:

Using $\log A + \log B = \log(AB)$:

4. Simplify the product. $(1+\cos x)(1-\cos x) = 1 - \cos^2 x = \sin^2 x$. So

And $\log(\sin^2 x) = 2\log|\sin x|$. Since $\sin x \ge 0$ on $[0,\pi]$, we can drop the absolute value:

Cancel the factor of 2:

5. Evaluate the standard integral $\int_{0}^{\pi} \log(\sin x)\,dx$. This is a classic result. One elegant way uses the same symmetry trick again. Let

Replace $x$ by $\pi - x$: since $\sin(\pi - x) = \sin x$, we get $J = J$ — that doesn't help directly. Instead, use the substitution $x = 2t$:

$J = \int_{0}^{\pi/2} \log(\sin 2t) \cdot 2\,dt = 2\int_{0}^{\pi/2} \log(2\sin t \cos t)\,dt.$ …

Method: Reflection + product-to-$\sin^2$, reducing to the standard $\int_0^\pi\log\sin x$

For $\displaystyle\int_0^\pi \log(1+\cos x)\,dx$-type integrals, reflect to get the complementary log, add so the product becomes $\sin^2x$, and use the known value of $\int_0^\pi\log\sin x\,dx$.

Steps

Step 1: Reflect $x\to\pi-x$.

Since $\cos(\pi-x)=-\cos x$, the reflected integrand is $\log(1-\cos x)$.

Step 2: Add the two forms and use $\log A+\log B=\log AB$. …

Common Mistakes

Mistake 1: Trying to find an antiderivative of $\log(1+\cos x)$.

Why it's wrong: it has no elementary antiderivative; reflection plus the product identity is the intended path. Correct approach: reflect $x\to\pi-x$ to get $\log(1-\cos x)$, then add.

Mistake 2: Forgetting $\log(\sin^2x)=2\log\sin x$ (valid since $\sin x\ge0$ on $[0,\pi]$). …