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Question: $\int \frac{sin^3x + cos^3x}{sin^2x \cdot cos^2x} dx$...

∫sin3x+cos3xsin2x⋅cos2xdx\int \frac{sin^3x + cos^3x}{sin^2x \cdot cos^2x} dx

Answer

sec⁡x−csc⁡x+C\sec x - \csc x + C

Explanation

Solution

The problem asks us to evaluate the integral ∫sin⁡3x+cos⁡3xsin⁡2x⋅cos⁡2xdx\int \frac{\sin^3x + \cos^3x}{\sin^2x \cdot \cos^2x} dx.

We can split the integrand into two separate fractions: ∫(sin⁡3xsin⁡2x⋅cos⁡2x+cos⁡3xsin⁡2x⋅cos⁡2x)dx\int \left( \frac{\sin^3x}{\sin^2x \cdot \cos^2x} + \frac{\cos^3x}{\sin^2x \cdot \cos^2x} \right) dx

Now, simplify each term: For the first term: sin⁡3xsin⁡2x⋅cos⁡2x=sin⁡xcos⁡2x\frac{\sin^3x}{\sin^2x \cdot \cos^2x} = \frac{\sin x}{\cos^2x} We can rewrite this as: sin⁡xcos⁡x⋅1cos⁡x=tan⁡x⋅sec⁡x\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \cdot \sec x

For the second term: cos⁡3xsin⁡2x⋅cos⁡2x=cos⁡xsin⁡2x\frac{\cos^3x}{\sin^2x \cdot \cos^2x} = \frac{\cos x}{\sin^2x} We can rewrite this as: cos⁡xsin⁡x⋅1sin⁡x=cot⁡x⋅csc⁡x\frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \cdot \csc x

Substitute these simplified terms back into the integral: ∫(tan⁡x⋅sec⁡x+cot⁡x⋅csc⁡x)dx\int (\tan x \cdot \sec x + \cot x \cdot \csc x) dx

Now, integrate each term separately using standard integral formulas: We know that: ∫sec⁡xtan⁡x dx=sec⁡x+C1\int \sec x \tan x \ dx = \sec x + C_1 And: ∫csc⁡xcot⁡x dx=−csc⁡x+C2\int \csc x \cot x \ dx = -\csc x + C_2

Combining these results, the integral becomes: ∫(tan⁡x⋅sec⁡x+cot⁡x⋅csc⁡x)dx=sec⁡x−csc⁡x+C\int (\tan x \cdot \sec x + \cot x \cdot \csc x) dx = \sec x - \csc x + C where CC is the constant of integration.

Explanation of the solution:

  1. Decomposition: The given fraction is split into two simpler fractions by distributing the denominator. sin⁡3x+cos⁡3xsin⁡2x⋅cos⁡2x=sin⁡3xsin⁡2x⋅cos⁡2x+cos⁡3xsin⁡2x⋅cos⁡2x\frac{\sin^3x + \cos^3x}{\sin^2x \cdot \cos^2x} = \frac{\sin^3x}{\sin^2x \cdot \cos^2x} + \frac{\cos^3x}{\sin^2x \cdot \cos^2x}
  2. Simplification: Each fraction is simplified using trigonometric identities:
    • The first term simplifies to sin⁡xcos⁡2x=sin⁡xcos⁡x⋅1cos⁡x=tan⁡xsec⁡x\frac{\sin x}{\cos^2x} = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x.
    • The second term simplifies to cos⁡xsin⁡2x=cos⁡xsin⁡x⋅1sin⁡x=cot⁡xcsc⁡x\frac{\cos x}{\sin^2x} = \frac{\cos x}{\sin x} \cdot \frac{1}{\sin x} = \cot x \csc x.
  3. Integration: The integral is then evaluated term by term using standard integration formulas:
    • ∫sec⁡xtan⁡x dx=sec⁡x\int \sec x \tan x \, dx = \sec x
    • ∫csc⁡xcot⁡x dx=−csc⁡x\int \csc x \cot x \, dx = -\csc x
  4. Final Result: Combining the results gives the final answer sec⁡x−csc⁡x+C\sec x - \csc x + C.