Question

If $\int e^{\sin x} \cdot [\frac{x\cos^3x-\sin x}{\cos^2x}] dx = e^{\sin x} f(x) + c$ where c is constant of integration, then $f(x)$ is equal to

• A. sec $x - x$
• B. $x -$sec $x$
• C. tan $x - x$
• D. $x -$ tan $x$

Answer 1

Answer: (B)

$x - \text{sec } x$

Answer 2

The given integral is $\int e^{\sin x} \cdot [\frac{x\cos^3x-\sin x}{\cos^2x}] dx = e^{\sin x} f(x) + c$.
Let's simplify the expression inside the square brackets:
$\frac{x\cos^3x-\sin x}{\cos^2x} = \frac{x\cos^3x}{\cos^2x} - \frac{\sin x}{\cos^2x} = x\cos x - \frac{\sin x}{\cos^2x} = x\cos x - \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = x\cos x - \tan x \sec x$.
So the integral becomes $\int e^{\sin x} (x\cos x - \tan x \sec x) dx$.

We are given that this integral is equal to $e^{\sin x} f(x) + c$.
Differentiating both sides with respect to $x$, we get:
$\frac{d}{dx} \left( \int e^{\sin x} (x\cos x - \tan x \sec x) dx \right) = \frac{d}{dx} (e^{\sin x} f(x) + c)$
The left side is the integrand:
$e^{\sin x} (x\cos x - \tan x \sec x)$
The right side is the derivative of the product $e^{\sin x} f(x)$:
$\frac{d}{dx} (e^{\sin x} f(x)) = e^{\sin x} \frac{d}{dx}(f(x)) + f(x) \frac{d}{dx}(e^{\sin x}) = e^{\sin x} f'(x) + f(x) e^{\sin x} \cos x = e^{\sin x} [f'(x) + f(x) \cos x]$.

Equating the left and right sides:
$e^{\sin x} (x\cos x - \tan x \sec x) = e^{\sin x} [f'(x) + f(x) \cos x]$
Dividing by $e^{\sin x}$ (which is always non-zero):
$x\cos x - \tan x \sec x = f'(x) + f(x) \cos x$
$f'(x) + (\cos x) f(x) = x\cos x - \tan x \sec x$

We need to find the function $f(x)$ that satisfies this equation. We can test the given options.
Option 1: $f(x) = \sec x - x$
$f'(x) = \sec x \tan x - 1$
$f'(x) + f(x) \cos x = (\sec x \tan x - 1) + (\sec x - x) \cos x = \sec x \tan x - 1 + \sec x \cos x - x\cos x = \sec x \tan x - 1 + 1 - x\cos x = \sec x \tan x - x\cos x$.
This is not equal to $x\cos x - \tan x \sec x$.

Option 2: $f(x) = x - \sec x$
$f'(x) = 1 - \sec x \tan x$
$f'(x) + f(x) \cos x = (1 - \sec x \tan x) + (x - \sec x) \cos x = 1 - \sec x \tan x + x\cos x - \sec x \cos x = 1 - \sec x \tan x + x\cos x - 1 = x\cos x - \sec x \tan x$.
This matches the required expression $x\cos x - \tan x \sec x$.
So, $f(x) = x - \sec x$ is the correct function.

Let's verify the general formula for integration: $\int e^{g(x)} [g'(x) f(x) + f'(x)] dx = e^{g(x)} f(x) + c$.
In this case, $g(x) = \sin x$, so $g'(x) = \cos x$.
The integrand is $e^{\sin x} (x\cos x - \tan x \sec x)$.
We need to check if $x\cos x - \tan x \sec x$ can be written in the form $f(x) \cos x + f'(x)$ for some function $f(x)$.
If we take $f(x) = x - \sec x$, then $f'(x) = 1 - \sec x \tan x$.
$f(x) \cos x + f'(x) = (x - \sec x)\cos x + (1 - \sec x \tan x) = x\cos x - \sec x \cos x + 1 - \sec x \tan x = x\cos x - 1 + 1 - \sec x \tan x = x\cos x - \sec x \tan x$.
This matches the term multiplying $e^{\sin x}$ in the integrand.
Thus, the integral is $\int e^{\sin x} [\cos x (x - \sec x) + (1 - \sec x \tan x)] dx = \int e^{\sin x} [g'(x) f(x) + f'(x)] dx$, where $g(x) = \sin x$ and $f(x) = x - \sec x$.
Using the formula, the integral is $e^{\sin x} f(x) + c = e^{\sin x} (x - \sec x) + c$.
Comparing this with the given equation, we find $f(x) = x - \sec x$.