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AP EAMCET · Maths · Definite Integration

If the value of 0π/2sin4x·cos2xdx=π32 then the value of 0π/2cos4x·sin2xdx=

  1. A π32
  2. B π64
  3. C π4
  4. D π8
Verified Solution

Answer & Solution

Correct Answer

(A) π32

Step-by-step Solution

Detailed explanation

Given, ∫0π2sin4x·cos2xdx=π32 ⇒∫0π2sin4π2-x·cos2π2-xdx=π32 ∵∫abfxdx=∫abfa+b-xdx ⇒∫0π2cos4x·sin2xdx=π32