Question

# ∫dxcotx2cotx3cotx6 =

A
lnsecx2secx3secx6+c
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B
lnsec2x2sec3x3sec6x6+c
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C
lnsec2x2cos3x3cos6x6+c
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D
lntan2x2tan3x3sec5x6+c
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Solution

## The correct option is C ln∣∣∣sec2x2cos3x3cos6x6∣∣∣+c∫dxcotx2cotx3cotx6I=∫tanx2tanx3tanx6dn⇒tanx6=tan(x2−x3) [tan(AB)=tanA−tanB1+tanA+tanB]tanx6=tanx/2+tanx/31+tanx/2tanx/3⇒tanx6(1+tanx2tanx3)=tanx2−tanx3⇒tanx6+tanx2tanx3tanx6=tanx2−tanx3Integrating both sides we gettanx6dx+∫tanx2tanx3tanx6dn=∫tanx2dx−∫tanx3dnI=∫tanx2dx−∫tanx3dn−∫tanx6dn [from equation]=2log∣∣∣secx2∣∣∣−3log∣∣∣secx3∣∣∣−6log∣∣∣secx6∣∣∣+C=log∣∣∣sec2x2∣∣∣+log∣∣∣cos3x3∣∣∣+log∣∣∣cos6x6∣∣∣+CI=log∣∣∣sec2x2+cos3x3+cos6x6∣∣∣+C

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