Q.Integrating factor of the differential equation $\cos x\frac{dy}{dx}+y\sin x=1$ is:
(A) $\cos x$
(B) $\tan x$
(C) $\sec x$
(D) $\sin x$

The key idea is to rewrite the equation in the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$ and then compute the integrating factor $\mu = e^{\int P \, dx}$.

First, divide through by $\cos x$:

$\frac{dy}{dx} + y \frac{\sin x}{\cos x} = \frac{1}{\cos x} \quad\Rightarrow\quad \frac{dy}{dx} + y \tan x = \sec x.$ …

The key to solving a first-order linear differential equation like

$\cos x \frac{dy}{dx} + y \sin x = 1$ is to recognise its structure.

A linear equation in $y$ has the form $\frac{dy}{dx} + P(x)y = Q(x)$.

The integrating factor (I.F.) is $e^{\int P\,dx}$ — it turns the left side into the derivative of $(y \cdot \text{I.F.})$, making the equation directly integrable.

Why does this work?

If you multiply the standard form by $e^{\int P\,dx}$, the left side becomes $\frac{d}{dx}\big(y e^{\int P\,dx}\big)$ by the product rule.

So the entire method hinges on correctly identifying $P(x)$.

1. Rewrite the equation in standard form The given equation is

Divide through by $\cos x$ (valid where $\cos x \neq 0$):

That is:

So here $P(x) = \tan x$ and $Q(x) = \sec x$.

2. Find the integrating factor The integrating factor is

Recall that $\int \tan x\,dx = \log|\sec x| + C$ (since $\tan x = \frac{\sin x}{\cos x}$ and the derivative of $\cos x$ is $-\sin x$, giving a natural log).

So

For the integrating factor, we usually take the simplest positive expression, so $\sec x$ (without the absolute value, assuming intervals where $\sec x > 0$). …

Method: Finding the Integrating Factor After Normalising the Leading Coefficient

To find the integrating factor of a linear equation whose $\frac{dy}{dx}$ has a coefficient other than $1$, divide first so that coefficient becomes $1$, then read $P(x)$.

Steps

Step 1: Divide through by the coefficient of $\frac{dy}{dx}$.

For $\cos x\frac{dy}{dx}+y\sin x=1$, divide by $\cos x$:

Step 2: Read $P(x)$ and integrate it.

Here $P(x)=\tan x$, and $\int\tan x\,dx=\log|\sec x|$. …

Common Mistakes

Mistake 1: Reading $P=\sin x$ before dividing by $\cos x$.

Why it's wrong: $P(x)$ is only valid once the coefficient of $\frac{dy}{dx}$ is $1$; dividing by $\cos x$ gives $P=\tan x$. Correct approach: normalise first, then $\text{I.F.}=e^{\int\tan x\,dx}=\sec x$.

Mistake 2: Evaluating $\int\tan x\,dx$ wrongly. …