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

Evaluate: ∫ (1+cos x)/(x+sin x)dx

NCERT Exemplar · Q5

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

Odisha ChseShort· 3mImportance★★★★★

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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 integral simplifies by noticing the numerator is the derivative of the denominator. Using the substitution $u = x + \sin x$ , the integral becomes $\int \frac{1}{u}\,du = \log|u| + C$ . The final answer is $\log|x + \sin x| + C$ .

The key insight here is pattern recognition: when you see an integral of the form $\int \frac{f'(x)}{f(x)}\,dx$ , the answer is always $\log|f(x)| + C$ . This is the natural logarithm rule in reverse — the derivative of $\log|f(x)|$ is $\frac{f'(x)}{f(x)}$ .

Let’s check if our integrand fits that pattern. The denominator is $x + \sin x$ . Its derivative is $1 + \cos x$ — which is exactly the numerator! That’s no coincidence; the problem was designed this way.

So we set $u = x + \sin x$ . Then $du = (1 + \cos x)\,dx$ , and the integral transforms cleanly.

Identify the substitution.

Let $u = x + \sin x$ . Then $du = (1 + \cos x)\,dx$ .
Rewrite the integral in terms of $u$ .

The original integral is $\int \frac{1 + \cos x}{x + \sin x}\,dx$ . Substituting, the numerator $1 + \cos x$ becomes $du/dx$ , and the denominator becomes $u$ :

\int \frac{1 + \cos x}{x + \sin x}\,dx = \int \frac{1}{u}\,du.

Integrate with respect to $u$ . The integral of $\frac{1}{u}$ is $\log|u| + C$ :

\int \frac{1}{u}\,du = \log|u| + C.

Substitute back for $x$ . Replace $u$ with $x + \sin x$ : …

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

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: Recognising $\int \frac{f'(x)}{f(x)}\,dx=\log|f(x)|+C$

Steps

Common Mistakes

Q.Evaluate: ∫1+cos⁡xx+sin⁡x dx\int \dfrac{1+\cos x}{x+\sin x}\,dx∫x+sinx1+cosx​dx

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

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