Q.Integrate the following function: $e^x \left(\frac{1+\sin x}{1+\cos x}\right)$

The key idea is to rewrite the integrand into a form that matches a known derivative — specifically, the derivative of $e^x f(x)$ is $e^x (f(x) + f'(x))$.

We want to express $e^x \left(\frac{1+\sin x}{1+\cos x}\right)$ as $e^x (f(x) + f'(x))$.

First, simplify the fraction:

A useful trick: multiply numerator and denominator by $(1 - \cos x)$ or use half-angle identities. Using $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $1+\cos x = 2\cos^2\frac{x}{2}$, we get:

$\frac{1+\sin x}{1+\cos x} = \frac{1 + 2\sin\frac{x}{2}\cos\frac{x}{2}}{2\cos^2\frac{x}{2}} = \frac{1}{2\cos^2\frac{x}{2}} + \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{1}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$ …

Why This Approach Works

When you see an integral of the form $\int e^x \cdot (\text{something})\,dx$, your first instinct should be to check if that "something" can be expressed as $f(x) + f'(x)$. Why? Because there's a beautiful shortcut:

This is a direct consequence of the product rule: $\frac{d}{dx}[e^x f(x)] = e^x f(x) + e^x f'(x)$. So if your integrand matches that pattern, the answer is simply $e^x f(x)$.

Our integrand is $e^x \left(\frac{1+\sin x}{1+\cos x}\right)$. The challenge is to rewrite $\frac{1+\sin x}{1+\cos x}$ as $f(x) + f'(x)$ for some cleverly chosen $f(x)$.

Step-by-Step Solution

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

Recall the standard half-angle formulas:

• $1 + \cos x = 2\cos^2\frac{x}{2}$
• $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$
• $1 + \sin x = \left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2$ (this is less common but useful)

Let's verify that last one: $\left(\sin\frac{x}{2} + \cos\frac{x}{2}\right)^2 = \sin^2\frac{x}{2} + \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} = 1 + \sin x$. Perfect.

So:

2. Split the square into two terms.

Now expand:

3. Use the identity $\tan^2\theta = \sec^2\theta - 1$ to simplify.

The $-1$ and $+1$ cancel neatly. So:

4. Spot the $f(x) + f'(x)$ pattern.

Let $f(x) = \tan\frac{x}{2}$. Then: …

Method: The $e^{x}\big(f(x)+f'(x)\big)$ shortcut

Whenever an integrand is $e^{x}$ times a bracket that is a function plus its own derivative, the answer is simply $e^{x}$ times that function:

Steps

Step 1: Isolate the $e^{x}$ factor and look at what multiplies it.

Step 2: Try to split the multiplier as $f(x)+f'(x)$.

Guess a candidate $f(x)$ (often the "nicer" of the two pieces, or the part left after a $\tfrac1x$-type term), then check that its derivative supplies the remaining piece. …

Common Mistakes

Mistake 1: Trying to integrate $\dfrac{1+\sin x}{1+\cos x}$ without simplifying.

Why it's wrong: it looks intractable until you use half-angle identities $1+\cos x=2\cos^{2}\tfrac x2$ and $1+\sin x=(\cos\tfrac x2+\sin\tfrac x2)^{2}$. Correct approach: simplify to $\tfrac12\sec^{2}\tfrac x2+\tan\tfrac x2$, which is $f+f'$ with $f=\tan\tfrac x2$.

Mistake 2: Differentiating $\tan\tfrac x2$ without the inner $\tfrac12$. …