Question
∫sin3x(cos4x+3cos2x+1)tan−1(secx+cosx)dx=
Answer: Option B
:
B
Put tan−1(secx+cosx)=f(x)⇒f′(x)=sin3xcos4x+3cos2x+1∴∫f′(x)f(x)dx=log|f(x)|+c∴Integral=log|tan−1(secx+cosx)|+c
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B
Put tan−1(secx+cosx)=f(x)⇒f′(x)=sin3xcos4x+3cos2x+1∴∫f′(x)f(x)dx=log|f(x)|+c∴Integral=log|tan−1(secx+cosx)|+c
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