∫ (sin x)/(3+4cos² x)dx = ____.

NCERT Exemplar · Q51

\int \dfrac{\sin x}{3+4\cos^2 x}\,dx =

Odisha ChseShort· 1mImportance★★★★★

Appeared in past exams:KCET 2024· Set A-1· 1mexactMHT-CET 2023· Set pcm-2023-05-09-E· 2mreworded

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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 rewrite the integral using a substitution that turns the denominator into a form matching the standard arctangent integral. The final result is $\boxed{-\frac{1}{2\sqrt{3}}\tan^{-1}\left(\frac{2\cos x}{\sqrt{3}}\right) + C}$ .

Why This Approach Works

When you see a denominator like $3 + 4\cos^2 x$ , your first instinct might be to try trigonometric identities. But notice the numerator is $\sin x$ , which is almost the derivative of $\cos x$ (up to a sign). That’s the real clue: whenever you have a function and its derivative nearby, substitution is the natural path.

The denominator is a sum of a constant and a square of $\cos x$ . After substitution, this becomes $a^2 + u^2$ , which is the classic form for the arctangent integral:

\int \frac{du}{a^2 + u^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C

So the plan is: let $u = \cos x$ , handle the sign from $du = -\sin x\,dx$ , and then match the constants.

Step-by-Step Solution

1. Choose the substitution.

Let $u = \cos x$ . Then $du = -\sin x\,dx$ , which means $\sin x\,dx = -du$ .

2. Rewrite the integral.

The original integral becomes:

\int \frac{\sin x}{3 + 4\cos^2 x}\,dx = \int \frac{-du}{3 + 4u^2}

3. Factor the denominator to match the standard form.

We want something like $\frac{1}{a^2 + u^2}$ . Factor out the 4:

3 + 4u^2 = 4\left(\frac{3}{4} + u^2\right) = 4\left(u^2 + \frac{3}{4}\right)

So the integral is:

\int \frac{-du}{4\left(u^2 + \frac{3}{4}\right)} = -\frac{1}{4} \int \frac{du}{u^2 + \frac{3}{4}}

4. Identify $a^2$ and apply the arctangent formula.

Here $a^2 = \frac{3}{4}$ , so $a = \frac{\sqrt{3}}{2}$ . Using $\int \frac{du}{u^2 + a^2} = \frac{1}{a} \tan^{-1}\left(\frac{u}{a}\right) + C$ :

$-\frac{1}{4} \cdot \frac{1}{\frac{\sqrt{3}}{2}} \tan^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right) + C = -\frac{1}{4} \cdot \frac{2}{\sqrt{3}} \tan^{-1}\left(\frac{2u}{\sqrt{3}}\right) + C$ …

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Q. $\int \dfrac{\sin x}{3+4\cos^2 x}\,dx =$ _______.

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?

Why This Approach Works

Step-by-Step Solution

Method: Substituting the trig function inside the denominator

Steps

Common Mistakes

Q.∫sin⁡x3+4cos⁡2x dx=\int \dfrac{\sin x}{3+4\cos^2 x}\,dx = ∫3+4cos2xsinx​dx= _______.

Why This Approach Works

Step-by-Step Solution

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Q. $\int \dfrac{\sin x}{3+4\cos^2 x}\,dx =$ _______.

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