Question

\( \int \frac{x+\cos x}{1-\sin x} dx = \)

• A. \( x \tan(\frac{\pi}{4}+\frac{x}{2}) + c \)
• B. \( x \tan(\frac{x}{2}) + c \)
• C. \( x \cot(\frac{x}{2}) + c \)
• D. \( x \cot(\frac{\pi}{4}+\frac{x}{2}) + c \)

Hint:

For tricky integrals, especially in multiple-choice questions, it's often faster to differentiate the given options than to perform the integration directly. If the derivative of an option matches the integrand, you've found the answer.

Answer 1

The Correct Option is A

Let's rewrite the integrand by splitting the fraction.
\( \int \left( \frac{x}{1-\sin x} + \frac{\cos x}{1-\sin x} \right) dx = \int \frac{x}{1-\sin x} dx + \int \frac{\cos x}{1-\sin x} dx \).
Let's try to use integration by parts on the first integral. Let \( u = x \) and \( dv = \frac{1}{1-\sin x} dx \).
To find v, we integrate dv: \( v = \int \frac{1}{1-\sin x} dx = \int \frac{1+\sin x}{1-\sin^2 x} dx = \int \frac{1+\sin x}{\cos^2 x} dx \).
\( v = \int (\sec^2 x + \tan x \sec x) dx = \tan x + \sec x \).
Now apply integration by parts \( \int u dv = uv - \int v du \):
\( \int \frac{x}{1-\sin x} dx = x(\tan x + \sec x) - \int (\tan x + \sec x) dx \).
So the original integral is \( [x(\tan x + \sec x) - \int (\tan x + \sec x) dx] + \int \frac{\cos x}{1-\sin x} dx \).
Let's focus on the last integral \( \int \frac{\cos x}{1-\sin x} dx \). Let \(t = 1-\sin x\), then \(dt = -\cos x dx\).
\( \int \frac{-dt}{t} = -\ln|t| = -\ln|1-\sin x| \). This doesn't seem to simplify things.
Let's try another approach. Consider the form \( \int (xf'(x)+f(x))dx = xf(x)+c \).
Let \(f(x) = \tan(\frac{\pi}{4}+\frac{x}{2})\).
\(f'(x) = \sec^2(\frac{\pi}{4}+\frac{x}{2}) \cdot \frac{1}{2}\).
We know that \( \tan(\frac{\pi}{4}+\frac{x}{2}) = \frac{1+\tan(x/2)}{1-\tan(x/2)} = \frac{\cos(x/2)+\sin(x/2)}{\cos(x/2)-\sin(x/2)} = \frac{1+\sin x}{\cos x} = \sec x + \tan x \).
And \( \sec^2(\frac{\pi}{4}+\frac{x}{2}) = 1 + \tan^2(\frac{\pi}{4}+\frac{x}{2}) = 1 + (\sec x + \tan x)^2\). This is getting complicated.
Let's go back to \(uv - \int v du\). The expression is \( x(\tan x + \sec x) - \int (\tan x + \sec x) dx + \int \frac{\cos x}{1-\sin x} dx \).
The integral \( \int \frac{\cos x}{1-\sin x} dx \) can be solved. Let's see if it cancels with \( \int (\tan x + \sec x) dx \). No.
Let's try to prove that \( \frac{d}{dx} [x \tan(\frac{\pi}{4}+\frac{x}{2})] = \frac{x+\cos x}{1-\sin x} \).
Let \( y = x \tan(\frac{\pi}{4}+\frac{x}{2}) \). Using product rule: \( \frac{dy}{dx} = 1 \cdot \tan(\frac{\pi}{4}+\frac{x}{2}) + x \cdot \sec^2(\frac{\pi}{4}+\frac{x}{2}) \cdot \frac{1}{2} \).
We know \( \tan(\frac{\pi}{4}+\frac{x}{2}) = \frac{1+\sin x}{\cos x} \). And \( \sec^2(\frac{\pi}{4}+\frac{x}{2}) = (\frac{1+\sin x}{\cos x})^2 \). No. Also, \( \sec^2\theta = \frac{1}{\cos^2\theta} \). \( \cos^2(\frac{\pi}{4}+\frac{x}{2}) = \frac{1+\cos(\pi/2+x)}{2} = \frac{1-\sin x}{2} \).
So, \( \sec^2(\frac{\pi}{4}+\frac{x}{2}) = \frac{2}{1-\sin x} \).
Substituting back: \( \frac{dy}{dx} = \frac{1+\sin x}{\cos x} + x \cdot \frac{2}{1-\sin x} \cdot \frac{1}{2} = \frac{1+\sin x}{\cos x} + \frac{x}{1-\sin x} \).
This is \( \frac{(1+\sin x)(1-\sin x) + x\cos x}{\cos x (1-\sin x)} = \frac{1-\sin^2 x + x \cos x}{\cos x(1-\sin x)} = \frac{\cos^2 x + x \cos x}{\cos x(1-\sin x)} = \frac{\cos x( \cos x + x)}{\cos x(1-\sin x)} = \frac{x+\cos x}{1-\sin x} \).
The derivative matches the integrand. Thus the integral is correct.