∫sin2xsin4x+cos4xdx is equal to tan−1(f(x)n)+C, then which of the following is/are correct ?
A
f(x)=tanx
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B
f(x)=cotx
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C
n=2
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D
n=4
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Solution
The correct option is Cn=2 I=∫sin2xsin4x+cos4xdx =∫2sinxcosxsin4x+cos4xdx =∫2tanxsec2x1+tan4xdx
Let tan2x=t⇒2tanxsec2xdx=dt ⇒I=∫dt1+t2=tan−1t+C=tan−1(tan2x)+C