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51.

`int(dx)/(sqrt(16-x^(2)))`

Answer» Correct Answer - `"sin"^(-1)(x)/(4)+C`
Discussion
52.

Evaluate: `int1/(sqrt(7-6x-x^2)) dx`

Answer» Correct Answer - `sin^(-1)((x+3)/(4))+C`
Discussion
53.

`int(dx)/(sqrt((x-3)^(2)-1))`

Answer» Correct Answer - `log|(x-3)+sqrt(x^(2)-6x+8)|+C`
Discussion
54.

`int (2x+3)/sqrt(x^2+x+1) dx`

Answer» Correct Answer - `2sqrt(x^(2)+x+1)+2log|x+(1)/(2)+sqrt(x^(2)+x+1)|+C`
Discussion
55.

Evaluate:`int(x^2)/(sqrt(1-x^6))dx`

Answer» Putting `x^(3)= t and x^(2) dx = (1)/(3)dt` , we get
`int(x^(2))/(sqrt(x^(6)-1))dx=(1)/(3)int(dt)/(sqrt(t^(2)-1))=(1)/(3)log|t+sqrt(t^(2)-1)|+C`
`=(1)/(3)log|x^(3)+sqrt(x^(6)-1)|+C`.
Discussion
56.

Evaluate : (i) `int(dx)/((9x^(2)-1))` (ii) `int(x)/((x^(4)-9))dx` (iii) `int(x^(2))/((x^(2)-9))dx`

Answer» We have (i) `int(dx)/((9x^(2)-1))=(1)/(9)*int(dx)/((x^(2)-(1)/(9)))`
`=(1)/(9)*int(dx)/({x^(2)-((1)/(3))^(2)})`
`=(1)/(9)*(1)/((2xx(1)/(3)))log|(x-(1)/(3))/(x+(1)/(3))|+C [becauseint(dx)/((x^(2)-a^(2)))=(1)/(2a)log|(x-a)/(x+a)|+C]`
`=(1)/(6)log|(3x-1)/(3x+1)|+C`.
(ii) Putting `x^(2)=t " and "2xdx=dt` , we get
`int(x)/((x^(4)-9))dx=(1)/(2)int(dt)/((t^(2)-9))=(1)/(2)*int(dt)/({t^(2)-(3)^(2)})`
`=(1)/(2)*(1)/((2xx3))log|(t-3)/(t+3)|+C`
`=(1)/(2)log|(x^(2)-3)/(x^(2)+3)|+C`.
(iii) `int(x^(2))/((x^(2)-9))dx=int{1+(9)/(x^(2)-9)}dx`
`=intdx+9int(dx)/({x^(2)-(3)^(2)})`
`=x+9*[(1)/((2xx3))log|(x-3)/(x+3)|]+C`
`=x+(3)/(2)log|(x-3)/(x+3)|+C`.
Discussion
57.

Evaluate: `int(5x+3)/(sqrt(x^2+4x+10)) dx`

Answer» Correct Answer - `5sqrt(x^(2)+4x+10)-7log|(x+2)+sqrt(x^(2)+4x+10)|+C`
Discussion
58.

Evaluate: `int(x+2)/(sqrt(x^2+2x-1)) dx`

Answer» Correct Answer - `sqrt(x^(2)+2x-1)+log|(x+1)+sqrt(x^(2)+2x-1)|+C`
Discussion
59.

`intsqrt(x^(2)+x)dx`

Answer» Correct Answer - `(1)/(4)(2x+1)sqrt(x^(2)+x)-(1)/(8)log|x+(1)/(2)++sqrt(x^(2)+x)|+C`
Discussion
60.

Evaluate : (i) `int(sinx)/(sqrt(4cos^(2)x-1))dx` (ii) `int(sec^(2)x)/(sqrt(tan^(2)x-4))dx`

Answer» (i) Putting `cosx=t and - sin x dx = dt` , we get
`int(sinx)/(sqrt(4 cos^(2)x-1))dx=int(-dt)/(sqrt(4t^(2)-1))=-(1)/(2)int(dt)/(sqrt(t^(2)-(1//2))^(2))`
`=-(1)/(2)*log|t+sqrt(t^(2)-(1)/(4))|+C`
`=-(1)/(2)*log|2t+sqrt(4t^(2)-1)|+C`
`=-(1)/(2)log|2cosx+sqrt(4cos^(2)x-1)|+C`.
Putting `tanx=t and sec^(2)x dx = dt` , we get
`int(sec^(2)x)/(sqrt(tan^(2)x-4))dx=int(dt)/(sqrt(t^(2)-4))=log|t+sqrt(t^(2)-4)|+C`
`=log|tanx+sqrt(tan^(2)x-4)|+C`.
Discussion
61.

`int (3x+1)/sqrt(5-2x-x^2) dx`

Answer» Correct Answer - `-3sqrt(5-2x-x^(2))-2sin^(-1)((x+1)/(sqrt(6)))+C`
Discussion
62.

Evaluate: `int(x+2) sqrt(x^2+x+1) dx`

Answer» Correct Answer - `(1)/(3)(x^(2)+x+1)^(3//2)+(3)/(8)(2x+1)sqrt(x^(2)+x+1)+(9)/(16)log|(x+(1)/(2))+sqrt(x^(2)+x+1)|+C`
Discussion
63.

Evaluate : (i) `int(sinx)/((1-4cos^(2)x))dx` (ii) `int("cosec"^(2)x)/((1-cot^(2)x))dx`

Answer» (i) Putting `cosx=t" and "-sinx dx=dt`, we get
`int(sinx)/((1-4cos^(2)x))dx=-int(dt)/((1-4t^(2)))`
`=-(1)/(4)*int(dt)/(((1)/(4)-t^(2)))=(1)/(4)*int(dt)/({((1)/(2))^(2)-t^(2)})`
`=-(1)/(4)xx(1)/((2xx(1)/(2)))log|((1)/(2)+t)/((1)/(2)-t)|+C`
`=-(1)/(4)log|(1+2t)/(1-2t)|+C=-(1)/(4)log|(1+2cosx)/(1-cosx)|+C`.
(ii) Putting `cotx=t" and -cosec"^(2)xdx=dt` we get
`int("cosec"^(2)x)/((1-cot^(2)x))dx=-int(dt)/((1-t^(2)))=-(1)/((2xx1))log|(1+t)/(1-t)|+C`
`=-(1)/(2)log|(1+cotx)/(1-cotx)|+C`.
Discussion
64.

`int(sinx)/(sqrt(4+cos^(2)x))dx`

Answer» Correct Answer - `-log|cosx+sqrt(4+cos^(2)x)|+C`
Put `cosx= t and sinx dx =-dt`.
Discussion
65.

`int(dx)/((sinx-2cosx)(2sinx+cosx))`

Answer» Correct Answer - `(1)/(5)log|(tanx-2)/(2tanx+1)|+C`
`I=int(dx)/((2sin^(2)x-3sinxcosx-2cos^(2)x))`.
Now , divide num . And denom , by `cos^(2)` x and put tan x =t.
Discussion