Integrate the following function: $ \frac{1}{1 - \tan x} $

Integrate the following function: (1)/(1 - tan x) – Class 12 Mathematics

Exercise 7.2 · Q33

\frac{1}{1 - \tan 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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Rewrite $\dfrac{1}{1-\tan x}$ as $\dfrac{\cos x}{\cos x-\sin x}$ , split $\cos x$ into half the sum of $(\cos x-\sin x)$ and $(\cos x+\sin x)$ , and integrate the two easy pieces. Final answer: $\dfrac{x}{2}-\dfrac12\log|\cos x-\sin x|+C$ .

Step 1 — Write in sines and cosines

Using $\tan x=\dfrac{\sin x}{\cos x}$ ,

\frac{1}{1-\tan x} = \frac{1}{1-\frac{\sin x}{\cos x}} = \frac{\cos x}{\cos x-\sin x}.

So we must find $I=\displaystyle\int \frac{\cos x}{\cos x-\sin x}\,dx$ .

Step 2 — The splitting trick

We want the numerator expressed through the denominator $\cos x-\sin x$ and its "partner" $\cos x+\sin x$ (whose combination gives the derivative of the denominator). Notice

(\cos x-\sin x)+(\cos x+\sin x)=2\cos x,

\cos x=\frac12\big[(\cos x-\sin x)+(\cos x+\sin x)\big].

Dividing by $\cos x-\sin x$ ,

\frac{\cos x}{\cos x-\sin x} = \frac12\left(1+\frac{\cos x+\sin x}{\cos x-\sin x}\right).

Step 3 — Integrate the two pieces

I=\frac12\int 1\,dx + \frac12\int \frac{\cos x+\sin x}{\cos x-\sin x}\,dx.

The first piece is $\dfrac{x}{2}$ .

For the second piece, put $u=\cos x-\sin x$ . Then

du=(-\sin x-\cos x)\,dx=-(\cos x+\sin x)\,dx,

so $(\cos x+\sin x)\,dx=-du$ and …

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Q.Integrate the following function: $\frac{1}{1 - \tan x}$

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?

Step 1 — Write in sines and cosines

Step 2 — The splitting trick

Step 3 — Integrate the two pieces

Method: Decompose numerator into denominator plus its derivative

Steps

Common Mistakes

Q.Integrate the following function: 11−tan⁡x\frac{1}{1 - \tan x}1−tanx1​

Step 1 — Write in sines and cosines

Step 2 — The splitting trick

Step 3 — Integrate the two pieces

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Q.Integrate the following function: $\frac{1}{1 - \tan x}$

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