Q.$\int_{\pi/3}^{\pi/2} \dfrac{\sqrt{1+\cos x}}{(1-\cos x)^{5/2}}\,dx$

The key idea is to simplify the integrand using half-angle identities, then evaluate using a standard trigonometric integral.

First, use the identities $1+\cos x = 2\cos^2\frac{x}{2}$ and $1-\cos x = 2\sin^2\frac{x}{2}$. Over the interval $[\pi/3, \pi/2]$, $\sin\frac{x}{2} > 0$ and $\cos\frac{x}{2} > 0$, so square roots are positive.

The integrand becomes:

Now substitute $u = \sin\frac{x}{2}$, so $du = \frac{1}{2}\cos\frac{x}{2}\,dx$, i.e. $\cos\frac{x}{2}\,dx = 2\,du$. When $x = \pi/3$, $u = \sin(\pi/6) = 1/2$; when $x = \pi/2$, $u = \sin(\pi/4) = 1/\sqrt{2}$.

The integral becomes: …

1. Apply half-angle identities. With $1+\cos x = 2\cos^2\frac x2$ and $1-\cos x = 2\sin^2\frac x2$ (both cosine terms positive on $[\pi/3,\pi/2]$),

So the integrand is

2. Substitute $u=\sin\frac x2$. Then $\cos\frac x2\,dx = 2\,du$, with $u:\tfrac12\to\tfrac{1}{\sqrt2}$ as $x:\tfrac{\pi}{3}\to\tfrac{\pi}{2}$:

$I = \frac14\int \frac{2\,du}{u^5} = \frac12\int_{1/2}^{1/\sqrt2} u^{-5}\,du.$ …

Method: Half-angle identities plus a power substitution

For definite integrals containing $\sqrt{1\pm\cos x}$ over $(1\mp\cos x)$ powers, convert everything to $\sin\tfrac x2$ and $\cos\tfrac x2$, then substitute.

Steps

Step 1: Apply the half-angle identities.

$1+\cos x=2\cos^2\tfrac x2$ and $1-\cos x=2\sin^2\tfrac x2$. On an interval where both half-angle terms are positive, the square roots open without a modulus.

Step 2: Reduce to a single trig ratio. …

Common Mistakes

Mistake 1: Ignoring the sign of the half-angle terms.

Why it's wrong: $\sqrt{2\cos^2\frac x2}=\sqrt2\,|\cos\frac x2|$ in general. Correct approach: check that on $[\pi/3,\pi/2]$ both $\sin\frac x2,\cos\frac x2>0$, so the roots are simply $\sqrt2\cos\frac x2$ and $2^{5/2}\sin^5\frac x2$.

Mistake 2: Botching the negative-power evaluation. …