Question
∫dxsin4x+cos4x is equal to
Answer: Option A
:
A
∫dxsin4x+cos4x=∫dx(sin2x+cos2x)2−2sin2xcos2x=∫2dx2−sin22x=∫2sec22x2sec22x−tan22xdx=2∫sec22x2+tan22xdx=∫dt(√2)2+t2[puttingtan2x=t⇒2sec22xdx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan2x)+C
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A
∫dxsin4x+cos4x=∫dx(sin2x+cos2x)2−2sin2xcos2x=∫2dx2−sin22x=∫2sec22x2sec22x−tan22xdx=2∫sec22x2+tan22xdx=∫dt(√2)2+t2[puttingtan2x=t⇒2sec22xdx=dt]=1√2tan−1(t√2)+C=1√2tan−1(1√2tan2x)+C
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