Q.Integrate the function $\frac{2+\sin 2x}{1+\cos 2x}\,e^x$

The key idea is to simplify the integrand using trigonometric identities, then apply the Power Rule for integration (reverse chain rule) after noticing a derivative pattern.

First, use $\sin 2x = 2\sin x \cos x$ and $\cos 2x = 2\cos^2 x - 1$, so $1 + \cos 2x = 2\cos^2 x$.

The integrand becomes:

$\frac{2 + 2\sin x \cos x}{2\cos^2 x}\,e^x = \frac{1 + \sin x \cos x}{\cos^2 x}\,e^x = (\sec^2 x + \tan x)\,e^x.$ …

We start with the integral

The presence of $e^x$ alongside trigonometric functions often hints at the pattern $\int e^x [f(x) + f'(x)]\,dx = e^x f(x) + C$. So our goal is to see if the given expression can be written in that form.

1. Simplify the denominator using the identity $\cos 2x = 2\cos^2 x - 1$.

   Then $1 + \cos 2x = 1 + (2\cos^2 x - 1) = 2\cos^2 x$.

2. Simplify the numerator using $\sin 2x = 2\sin x \cos x$.

   So $2 + \sin 2x = 2 + 2\sin x \cos x = 2(1 + \sin x \cos x)$.

3. Rewrite the fraction:

4. Split into two terms:

Because $\frac{\sin x}{\cos x} = \tan x$, the second term becomes $\tan x$.

5. The integral becomes:

6. Recognise the derivative pattern: Let $f(x) = \tan x$. Then $f'(x) = \sec^2 x$. So the integrand is exactly $e^x [f(x) + f'(x)]$. …

Method: Recognise the pattern $\int e^{x}\,[f(x)+f'(x)]\,dx=e^{x}f(x)$

Use this whenever $e^{x}$ multiplies a trig/algebraic expression: simplify the bracket and check whether it is some $f$ plus its own derivative.

Steps

Step 1: Simplify the non-exponential factor.

Use double-angle identities ($1+\cos 2x=2\cos^2x$, $\sin 2x=2\sin x\cos x$) to reduce the messy ratio to a clean trig expression such as $\sec^2x+\tan x$.

Step 2: Split into $f+f'$. …

Common Mistakes

Mistake 1: Reaching for integration by parts.

Why it's wrong: parts on $e^x\sec^2x$ and $e^x\tan x$ separately is long and error-prone; the $f+f'$ pattern gives the answer in one line. Correct approach: recognise $\int e^x[f+f']\,dx=e^x f$.

Mistake 2: Errors in the double-angle simplification. …