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Evaluate:`int((cos2x)^(1/2))/(sinx)dx` |
Answer» Correct Answer - `-"log"|"cot x" + sqrt("cot"^(2)x-1|)+(1)/(sqrt(2))"log"|(sqrt(2)+sqrt(1- "tan"^(2)x))/(sqrt(2)-sqrt(1- "tan"^(2)x))|+c` Let `I=int(sqrt(cos2x))/(sinx)dx=int sqrt((cos^(2)x-sin^(2)x)/(sin^(2)x))dx` `=intsqrt(cot^(2)x-1)dx` Put `cot x = sec theta rArr - "cosec"^(2)x dx = sec theta tan theta d theta` `therefore I=int sqrt(sec^(2)theta-1)*(sec theta*tan theta)/(-(1+sec^(2)theta))d theta` `=-int(sec theta*tan^(2)theta)/(1+ sec^(2)theta)d theta` `=-int (sin^(2)theta)/(cos theta + cos^(3)theta)d theta` `=-int(1-cos^(2)theta)/(cos theta(1+cos^(2)theta))d theta` `=-int((1+cos^(2)theta)-2cos^(2)theta)/(cos theta(1+cos^(2)theta))d theta` `=-intsec theta d theta +2 int(cos theta)/(1+cos^(2)theta)d theta` `=-log|sec theta+tan theta|+2 int(cos theta)/(2-sin^(2)theta)d theta` `=-log|sec theta+tan theta|+ int(dt)/(2-t^(2)), "where sin" theta=t` `=-log|sec theta + tan theta|+2*(1)/(2 sqrt(2))log|(sqrt(2)+sin theta)/(sqrt(2)-sin theta)|+c` `=-log|cot x +sqrt(cot^(2)x-1)|+(1)/(sqrt(2))log|(sqrt(2)+sqrt(1-tan^(2)x))/(sqrt(2)-sqrt(1-tan^(2)x))|+c` |
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