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a
123log3-tanx3+tanx+C
b
13log3+tanx3−tanx+C
c
123log3+tanx3−tanx+C
d
12log3+tanx3−tanx+C
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Correct option is C
I=∫sinxsin3xdx=∫sinx3sinx−4sin3xdx=∫13−4sin2xdx=∫sec2x3sec2x−4tan2xdx Dividing Nr and Dr by cos2x Putting tan x = t and sec2x dx = dt, we getI=∫dt31+t2−4t2=∫dt3−t2=∫1(3)2−t2dt=123log3+t3−t+C=123log3+tanx3−tanx+CSimilar Questions
is equal to
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