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Answer» Correct Answer - `a to p,q; b to r,s; c to p; d to p,q` a. `" Let " I=int(2^(x))/(sqrt(1-4^(x)))dx=(1)/(log2)int(1)/(sqrt(1^(2)-t^(2)))dt` `"Putting "2^(x)=t, 2^(x)log2dx=dt, " we get "` `I=(1)/(log2)sin^(-1)((t)/(1))+C=(1)/(log 2)sin^(-1)(2^(x))+C` ` :. k=(1)/(log2)` b. ` int(dx)/((sqrt(x))^(2)+(sqrt(x))^(7))=int(dx)/((sqrt(x))^(7)(1+(1)/((sqrt(x))^(5))))` `"Put " (1)/((sqrt(x))^(5))=y, (dy)/(dx)=-(5)/(2(sqrt(x))^(7))` ` :. I=int(-2dy)/(5(1+y))=-(2)/(5)In|1+y|+C=(2)/(5)In((1)/(1+(1)/((sqrt(x))^(5))))` ` :. a=(2)/(5), k=(5)/(2)` c. Add and subtract ` 2x^(2)` in the numerator. Then `k=1` and ` m=1 ` d. `I=int(dx)/(5+4cosx)` `=int(dx)/(5("sin"^(2)(x)/(2)+"cos"^(2)(x)/(2))+4("cos"^(2)(x)/(2)-"sin"^(2)(x)/(2)))` `=int(dx)/(9"cos"^(2)(x)/(2)+"sin"^(2)(x)/(2))=int("sec"^(2)(x)/(2))/(9+"tan"^(2)(x)/(2))dx` `"Let " t="tan"(x)/(2) " or " 2dt="sec"^(2)(x)/(2)dx` ` :. I=int(2dt)/(9+t^(2))=(2)/(3)"tan"^(-1)((t)/(3))+C` `=(2)/(3)"tan"^(-1)(("tan"((x)/(2)))/(3))+C` ` :. k=(2)/(3), m=(1)/(3)` |
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