Question

If \(k\in \mathbb{N}\), then \[ \int \frac{1-k\cos^2 x}{\sin^k x\cos^2 x}\,dx= \]

• A. \(\dfrac{\tan x}{\sin^{k+1}x}+C\)
• B. \(\dfrac{\tan x}{\sin^k x}+C\)
• C. \(\sin^k x\sec^2 x+C\)
• D. \(k\sin^{k-1}x\cos x+C\)

Hint:

In integration, if the integrand looks complicated, check whether it is the derivative of one of the options. Differentiating the options is often the fastest method in MCQs.

Answer 1

The Correct Option is B

Step 1: Observe the structure of the integrand.
The given integral is \[ \int \frac{1-k\cos^2 x}{\sin^k x\cos^2 x}\,dx \] We need to identify a function whose derivative gives this expression.
From the options, consider \[ \frac{\tan x}{\sin^k x} \]

Step 2: Rewrite the expression.
Let \[ y=\frac{\tan x}{\sin^k x} \] Since \[ \tan x=\frac{\sin x}{\cos x}, \] we get \[ y=\frac{\sin x}{\cos x\sin^k x} \] So, \[ y=\frac{1}{\cos x\sin^{k-1}x} \]

Step 3: Differentiate using product form.
Write \[ y=\sec x\sin^{1-k}x \] Now differentiate: \[ \frac{dy}{dx}=\sec x\tan x\sin^{1-k}x+\sec x(1-k)\sin^{-k}x\cos x \] Since \[ \sec x\cos x=1, \] we get \[ \frac{dy}{dx}=\sec x\tan x\sin^{1-k}x+(1-k)\sin^{-k}x \] Now, \[ \sec x\tan x=\frac{\sin x}{\cos^2 x} \] So, \[ \sec x\tan x\sin^{1-k}x = \frac{\sin x}{\cos^2 x}\cdot \sin^{1-k}x \] \[ =\frac{\sin^{2-k}x}{\cos^2 x} \] Also, \[ (1-k)\sin^{-k}x=\frac{1-k}{\sin^k x} \]

Step 4: Take common denominator.
Thus, \[ \frac{dy}{dx} = \frac{\sin^{2-k}x}{\cos^2 x}+\frac{1-k}{\sin^k x} \] Taking common denominator \(\sin^k x\cos^2 x\), \[ \frac{dy}{dx} = \frac{\sin^2 x+(1-k)\cos^2 x}{\sin^k x\cos^2 x} \] Using \[ \sin^2 x=1-\cos^2 x, \] we get \[ \frac{dy}{dx} = \frac{1-\cos^2 x+(1-k)\cos^2 x}{\sin^k x\cos^2 x} \] \[ \frac{dy}{dx} = \frac{1-k\cos^2 x}{\sin^k x\cos^2 x} \]

Step 5: Match with the integrand.
Therefore, \[ \frac{d}{dx}\left(\frac{\tan x}{\sin^k x}\right) = \frac{1-k\cos^2 x}{\sin^k x\cos^2 x} \] Hence, \[ \int \frac{1-k\cos^2 x}{\sin^k x\cos^2 x}\,dx = \frac{\tan x}{\sin^k x}+C \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\frac{\tan x}{\sin^k x}+C} \]