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Evaluate: `int(sinx)/(cos2x) dx` |
Answer» Correct Answer - ` -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C` `I=int(sinx)/(cos2x)dx=int(sinx)/(2cos^(2)x-1)dx` ` "Let " cosx=t` ` :. -sinx dx=dt` ` :. I=int(-dt)/(2t^(2)-1)= -(1)/(2)int(dt)/(t^(2)-(1)/(2))` `=-(1)/(2)*(1)/(2(1)/(sqrt(2)))"log"_(e)|(t-(1)/(sqrt(2)))/(t+(1)/(sqrt(2)))|+C` `= -(1)/(2sqrt(2))"log"_(e)|(sqrt(2)t-1)/(sqrt(2)t+1)|+C` |
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