Question
∫1−7 cos2xsin7xcos2xdx=f(x)(sinx)7+C, then f(x) is equal to
Answer: Option C
:
C
∫1−7cos2xsin7xcos2xdx=∫(sec2xsin7x−7sin7x)dx=∫sec2xsin7xdx−∫7sin7dx=I1+I2Now,I1=∫sec2xsin7dx=tanxsin7x+7∫tanxcosxsin8x=tanxsin7x−I2∴I1+I2=tanxsin7x+C⇒f(x)=tanx
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:
C
∫1−7cos2xsin7xcos2xdx=∫(sec2xsin7x−7sin7x)dx=∫sec2xsin7xdx−∫7sin7dx=I1+I2Now,I1=∫sec2xsin7dx=tanxsin7x+7∫tanxcosxsin8x=tanxsin7x−I2∴I1+I2=tanxsin7x+C⇒f(x)=tanx
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