Q.Evaluate the definite integral $\int_{\pi/2}^{\pi}e^x\left(\frac{1-\sin x}{1-\cos x}\right)dx$

We first simplify the integrand using trigonometric identities.

Step 1 – Rewrite using half-angle formulas

Recall:

$1 - \sin x = (\sin\frac{x}{2} - \cos\frac{x}{2})^2$

$1 - \cos x = 2\sin^2\frac{x}{2}$

Thus

Step 2 – Expand and simplify

Using $\cot^2\theta = \csc^2\theta - 1$, we get

So the integral becomes

Step 3 – Recognize derivative pattern

Notice that $\frac{d}{dx}\left(e^x \cot\frac{x}{2}\right) = e^x\cot\frac{x}{2} - \frac12 e^x\csc^2\frac{x}{2}$. …

Why This Approach Works

When you see an integral of the form $\int e^x f(x) \, dx$, your mind should immediately check whether $f(x) + f'(x)$ appears somewhere. That’s because the derivative of $e^x f(x)$ is $e^x [f(x) + f'(x)]$ — a neat property of the exponential function. If we can express the given integrand as $e^x$ times something that looks like $f(x) + f'(x)$, the integral collapses to $e^x f(x)$ plus a constant.

Here, the integrand is $e^x \cdot \frac{1-\sin x}{1-\cos x}$. The trigonometric part looks messy, but it’s actually a disguised form of something simpler. Let’s clean it up.

Step-by-Step Solution

1. Simplify the trigonometric fraction using half-angle identities.

Recall:

So:

Split the numerator:

The first term is $\frac{1}{2}\csc^2\frac{x}{2}$. The second term simplifies to $\cot\frac{x}{2}$ (since $\frac{\cos}{\sin} = \cot$). So:

2. Now the integral becomes:

3. Spot the derivative pattern.

Let’s guess a function $f(x)$ such that $f(x) + f'(x)$ equals the bracket. Try $f(x) = -\cot\frac{x}{2}$.

Compute $f'(x)$:

So:

which is exactly the bracket we have. …

Method: Half-angle simplification into the $e^{x}[f+f']$ pattern

Use this when $e^{x}$ multiplies a ratio like $\dfrac{1-\sin x}{1-\cos x}$: half-angle identities recast the bracket as $f+f'$ so the antiderivative is $e^x f$.

Steps

Step 1: Convert with half-angle identities.

Use $1-\cos x=2\sin^2\tfrac{x}{2}$ and $\sin x=2\sin\tfrac{x}{2}\cos\tfrac{x}{2}$ to rewrite the ratio as $\tfrac12\csc^2\tfrac{x}{2}-\cot\tfrac{x}{2}$.

Step 2: Match to $f(x)+f'(x)$. …

Common Mistakes

Mistake 1: Not using half-angle identities.

Why it's wrong: $\dfrac{1-\sin x}{1-\cos x}$ only reveals the $f+f'$ structure after converting to $\tfrac12\csc^2\tfrac x2-\cot\tfrac x2$. Correct approach: rewrite with $1-\cos x=2\sin^2\tfrac x2$.

Mistake 2: Forgetting the chain-rule $\tfrac12$ in $\tfrac{d}{dx}\cot\tfrac x2$.

Why it's wrong: $\tfrac{d}{dx}\bigl(-\cot\tfrac x2\bigr)=\tfrac12\csc^2\tfrac x2$; missing the $\tfrac12$ breaks the $f+f'$ match. Correct approach: differentiate the half-angle carefully. …