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a
23tan2x−3tanx+C
b
32tan2x−3tanx+C
c
−13tanx+3tan2x+C
d
−25tan2x−3tanx+C
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Correct option is A
I=∫dxsin3xcos5x Dividing Nr∈Dr by cos4x=∫sec4xsin3xcos5xcos8xdx=∫sec4xtan3xdx=∫sec2xsec2xtan2x⋅tanxdx=∫1+tan2xsec2xtanx⋅tanxdx Put tanx=tsec2xdx=dt=∫1+t2dttt=∫t−32+t12dt=t−32+1−32+1+t12+112+1+c=t−12−12+t3232+c=+2−1t⋅+tt3+c=2−3+t23t+c=23t2−3t+c=23tan2x−3tanx+cSimilar Questions
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