Evaluate: $\int \dfrac{dx}{1+\cos x}$

Evaluate: ∫ (dx)/(1+cos x)

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\int \dfrac{dx}{1+\cos x}

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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 key idea is to use the half-angle identity $1+\cos x = 2\cos^2(x/2)$ to rewrite the integral, then substitute $u = x/2$ to get a standard $\int \sec^2 u \, du$ form. The result is $\tan(x/2) + C$ .

When you see $1 + \cos x$ in the denominator of an integral, your first instinct might be to try a trigonometric identity. The expression $1 + \cos x$ is a classic signal for the half-angle formula. Why? Because $1 + \cos x$ simplifies beautifully to $2\cos^2(x/2)$ , which turns a messy rational function into something clean and integrable.

The intuition: we want to eliminate the sum $1 + \cos x$ because it’s not a standard derivative. But $\cos^2(x/2)$ is the square of a cosine, and its reciprocal is $\sec^2(x/2)$ , whose integral is $\tan(x/2)$ — a direct match. So the half-angle identity is the natural path.

Let’s work through it step by step.

Apply the half-angle identity. Recall: $\cos x = 2\cos^2(x/2) - 1$ , so $1 + \cos x = 2\cos^2(x/2)$ . This gives:

\int \frac{dx}{1+\cos x} = \int \frac{dx}{2\cos^2(x/2)} = \frac12 \int \sec^2\left(\frac{x}{2}\right) dx.

Substitute to match the standard form. Let $u = x/2$ , so $du = \frac12 dx$ , meaning $dx = 2\,du$ . The integral becomes:

\frac12 \int \sec^2(u) \cdot 2\,du = \int \sec^2 u \, du.

Integrate the secant-squared. The integral of $\sec^2 u$ is $\tan u + C$ . So: $\int \sec^2 u \, du = \tan u + C.$ …

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Q.Evaluate: $\int \dfrac{dx}{1+\cos x}$

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?

Method: Half-angle reduction for $\int\frac{dx}{1\pm\cos x}$ (and $1\pm\sin x$ )

Steps

Common Mistakes

Q.Evaluate: ∫dx1+cos⁡x\int \dfrac{dx}{1+\cos x}∫1+cosxdx​

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Q.Evaluate: $\int \dfrac{dx}{1+\cos x}$

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