Q.Choose the correct answer: $\int \frac{\cos 2x}{(\sin x+\cos x)^2}\,dx$ is equal to (A) $\frac{-1}{\sin x+\cos x}+C$ (B) $\log|\sin x+\cos x|+C$ (C) $\log|\sin x-\cos x|+C$ (D) $\frac{1}{(\sin x+\cos x)^2}$

Concept: U Substitution — simplify the denominator using a trigonometric identity.

First, note that

Also, $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$.

So the integral becomes

$\int \frac{\cos 2x}{(\sin x + \cos x)^2}\,dx = \int \frac{(\cos x - \sin x)(\cos x + \sin x)}{(\sin x + \cos x)^2}\,dx = \int \frac{\cos x - \sin x}{\sin x + \cos x}\,dx.$ …

When you first look at $\int \frac{\cos 2x}{(\sin x+\cos x)^2}\,dx$, the denominator is a square of a sum, and the numerator is $\cos 2x$. Your instinct might be to expand $\cos 2x$ as $\cos^2 x - \sin^2 x$ or $1 - 2\sin^2 x$, but that leads to messy algebra. The cleaner path is to notice that $\cos 2x$ factorises beautifully: $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$. This is the insight that unlocks the problem.

Why does this help? Because the denominator is $(\sin x + \cos x)^2$, so one factor of $(\sin x + \cos x)$ cancels with the same factor in the numerator. What remains is a fraction where the numerator is the derivative of the denominator (up to a sign), which screams for a $u$-substitution.

Let’s work through it step by step.

1. Rewrite the numerator $\cos 2x = \cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$. So the integral becomes

2. Cancel the common factor Since $\sin x + \cos x = \cos x + \sin x$, one factor cancels:

This is much simpler.

3. Spot the substitution Let $u = \sin x + \cos x$. Then $du = (\cos x - \sin x)\,dx$. …

Method: Factor $\cos 2x$ to cancel against $(\sin x+\cos x)^{2}$

Use this when $\cos 2x$ sits over $(\sin x\pm\cos x)^2$: factoring $\cos 2x=\cos^2x-\sin^2x$ produces a common factor that cancels, leaving a $\log$-type integral.

Steps

Step 1: Factor the numerator.

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

Step 2: Cancel with one factor of the denominator. …

Common Mistakes

Mistake 1: Expanding $(\sin x+\cos x)^2$ into $1+\sin 2x$.

Why it's wrong: while valid, it makes the integral messier; the clean route factors $\cos 2x$ to cancel a common factor. Correct approach: use $\cos 2x=(\cos x-\sin x)(\cos x+\sin x)$.

Mistake 2: Not factoring $\cos 2x$.

Why it's wrong: without factoring, the cancellation with $(\sin x+\cos x)^2$ is missed. Correct approach: factor and cancel one $(\sin x+\cos x)$.

Mistake 3: Sign error in $du$. …