Question

If \[ \int \frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx = A\sec x+B\cosec x+c, \] then \((A,B)\) are

• A. \((1,1)\)
• B. \((-1,-1)\)
• C. \((1,-1)\)
• D. \((-1,1)\)

Hint:

Try to convert integrands into known derivative forms like \(\sec x\tan x\) and \(\cosec x\cot x\).

Answer 1

The Correct Option is C

Concept: We split the fraction and use: \[ \frac{\sin x}{\cos^2x}=\tan x\sec x \] and \[ \frac{\cos x}{\sin^2x}=\cot x\cosec x \]

Step 1: Given integral: \[ \int \frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx \]

Step 2: Split the numerator. \[ \frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x} = \frac{\sin^3x}{\sin^2x\cos^2x} + \frac{\cos^3x}{\sin^2x\cos^2x} \] \[ = \frac{\sin x}{\cos^2x} + \frac{\cos x}{\sin^2x} \]

Step 3: Convert into standard derivative forms. \[ \frac{\sin x}{\cos^2x} = \tan x\sec x \] \[ \frac{\cos x}{\sin^2x} = \cot x\cosec x \] So, \[ \int \frac{\sin^3x+\cos^3x}{\sin^2x\cos^2x}\,dx = \int \left(\tan x\sec x+\cot x\cosec x\right)\,dx \]

Step 4: Use standard formulas. \[ \int \sec x\tan x\,dx=\sec x \] \[ \int \cosec x\cot x\,dx=-\cosec x \] Therefore, \[ \int \left(\tan x\sec x+\cot x\cosec x\right)\,dx = \sec x-\cosec x+c \]

Step 5: Compare with: \[ A\sec x+B\cosec x+c \] We get: \[ A=1,\qquad B=-1 \] Hence, \[ \boxed{(A,B)=(1,-1)} \]