Asked by MANDAKRANTA CHAKRABORTY | 20th Mar, 2012, 12:22: PM
In the integration of sinx / cos2x , we can write the expression as :
sinx/(2cos2x - 1) , then put cosx = t , sinx.dx = -dt
we get Integration of -dt/(2t2 - 1) = (-1/2) . (1/root2) . log[((root2).t - 1)/((root2).t + 1)]
then put t = cosx in the above expression .
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Answered by | 20th Mar, 2012, 01:27: PM
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