Integrate the following function: $\frac{\cos 2x}{(\cos x + \sin x)^2}$

Integrate the following function: (cos 2x)/((cos x + sin x)²)

Exercise 7.3 · Q20

\frac{\cos 2x}{(\cos x + \sin x)^2}

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given $\cos(x^2) \cdot 2x$ , find the original function. That's what u substitution does — it reverses the chain rule.

The Core Intuition

When an integral looks like "a function times the derivative of its inside," substitute the inside with $u$ and the derivative of the inside with $du$ . Consider:

\int 2x \cos(x^2) \, dx

Here $2x$ is the derivative of $x^2$ , and $x^2$ is the inside of $\cos(x^2)$ . Let $u = x^2$ , so $du = 2x \, dx$ :

\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of $\sin(x^2)$ is $\cos(x^2) \cdot 2x$ .

The Precise Statement

\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided $g$ is differentiable and the resulting integral in $u$ is simpler.

The Step-by-Step Method

Identify a function $g(x)$ whose derivative $g'(x)$ also appears (possibly up to a constant factor).
Set $u = g(x)$ , compute $du = g'(x) \, dx$ .
Rewrite the entire integral in $u$ and $du$ — every $x$ and $dx$ must be replaced.
Integrate with respect to $u$ .
Substitute back $u = g(x)$ .

Watch out

You cannot mix variables. If any $x$ remains after substitution, you chose the wrong $u$ (or must solve for $x$ in terms of $u$ — rare).

A Second Example (with a constant factor)

Evaluate $\int x \sqrt{x^2 + 1} \, dx$ . Let $u = x^2 + 1$ , so $x \, dx = \frac{1}{2} du$ :

\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

$x \cdot f(x^2)$ — derivative of $x^2$ is $2x$ , so $u = x^2$
$e^{g(x)} \cdot g'(x)$ — derivative of $g(x)$ appears
$\frac{g'(x)}{g(x)}$ — leads to $\log|g(x)|$ …

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The integrand simplifies to $\frac{\cos x-\sin x}{\cos x+\sin x}$ , whose integral is $\log\lvert\cos x+\sin x\rvert+C=\frac12\log\lvert 1+\sin 2x\rvert+C$ .

Simplify first

Expand the denominator: $(\cos x+\sin x)^2=\cos^2 x+\sin^2 x+2\sin x\cos x=1+\sin 2x$ . For the numerator, $\cos 2x=\cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x)$ . So one factor of $(\cos x+\sin x)$ cancels:

\frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^2}=\frac{\cos x-\sin x}{\cos x+\sin x}.

Substitute

The numerator is exactly the derivative of the denominator: with $u=\cos x+\sin x$ , $du=(-\sin x+\cos x)\,dx=(\cos x-\sin x)\,dx$ . Therefore

\int\frac{\cos x-\sin x}{\cos x+\sin x}\,dx=\int\frac{du}{u}=\log|u|+C=\log\lvert\cos x+\sin x\rvert+C.

Equivalent form …

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Q.Integrate the following function: $\frac{\cos 2x}{(\cos x + \sin x)^2}$

Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The Core Intuition

The Precise Statement

The Step-by-Step Method

A Second Example (with a constant factor)

When Does It Work?

Simplify first

Substitute

Equivalent form …

Method: Factor $\cos 2x$ to cancel a $(\cos x+\sin x)$ denominator

Steps

Common Mistakes

Q.Integrate the following function: cos⁡2x(cos⁡x+sin⁡x)2\frac{\cos 2x}{(\cos x + \sin x)^2}(cosx+sinx)2cos2x​

Simplify first

Substitute

Equivalent form …

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Q.Integrate the following function: $\frac{\cos 2x}{(\cos x + \sin x)^2}$

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