Question

The integral $\int (\frac{x}{x \sin x + \cos x})^2 dx$ is equal to: (2020)

(where C is a constant of integration):

• A. $\tan x - \frac{x \sec x}{x \sin x + \cos x} + C$
• B. $\tan x + \frac{x \sec x}{x \sin x + \cos x} + C$
• C. $\sec x - \frac{x \tan x}{x \sin x + \cos x} + C$
• D. $\sec x + \frac{x \tan x}{x \sin x + \cos x} + C$

Answer 1

Answer: (A)

$\tan x - \frac{x \sec x}{x \sin x + \cos x} + C$

Answer 2

To evaluate the integral $I = \int \left(\frac{x}{x \sin x + \cos x}\right)^2 dx$, we can use the method of integration by parts.

Let's denote the expression in the denominator as $f(x) = x \sin x + \cos x$. First, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx}(x \sin x + \cos x)$ Using the product rule for $x \sin x$: $\frac{d}{dx}(x \sin x) = 1 \cdot \sin x + x \cdot \cos x$. Using the derivative for $\cos x$: $\frac{d}{dx}(\cos x) = -\sin x$. So, $f'(x) = (\sin x + x \cos x) - \sin x = x \cos x$.

Now, rewrite the integrand in a form suitable for integration by parts. The integrand is $\frac{x^2}{(x \sin x + \cos x)^2}$. We can write this as $\frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x + \cos x)^2}$. This form suggests choosing: $u = \frac{x}{\cos x} = x \sec x$ $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} dx = \frac{f'(x)}{(f(x))^2} dx$

Now, we find $du$ and $v$: $du = \frac{d}{dx}(x \sec x) dx = (1 \cdot \sec x + x \cdot \sec x \tan x) dx = (\sec x + x \sec x \tan x) dx$. $v = \int \frac{f'(x)}{(f(x))^2} dx$. Let $t = f(x)$, then $dt = f'(x) dx$. So $v = \int \frac{1}{t^2} dt = -\frac{1}{t} = -\frac{1}{x \sin x + \cos x}$.

Apply the integration by parts formula: $\int u dv = uv - \int v du$. $I = (x \sec x) \left(-\frac{1}{x \sin x + \cos x}\right) - \int \left(-\frac{1}{x \sin x + \cos x}\right) (\sec x + x \sec x \tan x) dx$ $I = -\frac{x \sec x}{x \sin x + \cos x} + \int \frac{\sec x + x \sec x \tan x}{x \sin x + \cos x} dx$

Now, let's evaluate the remaining integral $J = \int \frac{\sec x + x \sec x \tan x}{x \sin x + \cos x} dx$. Simplify the numerator: $\sec x + x \sec x \tan x = \frac{1}{\cos x} + x \frac{1}{\cos x} \frac{\sin x}{\cos x}$ $= \frac{\cos x}{\cos^2 x} + \frac{x \sin x}{\cos^2 x} = \frac{\cos x + x \sin x}{\cos^2 x}$

Substitute this back into the integral $J$: $J = \int \frac{\frac{\cos x + x \sin x}{\cos^2 x}}{x \sin x + \cos x} dx$ Assuming $x \sin x + \cos x \neq 0$, we can cancel the term $(\cos x + x \sin x)$ from the numerator and denominator: $J = \int \frac{1}{\cos^2 x} dx = \int \sec^2 x dx$ $J = \tan x + C'$ (where $C'$ is the constant of integration for $J$)

Substitute $J$ back into the expression for $I$: $I = -\frac{x \sec x}{x \sin x + \cos x} + \tan x + C$

Rearranging the terms, we get: $I = \tan x - \frac{x \sec x}{x \sin x + \cos x} + C$

Comparing this result with the given options, our result matches option (A).