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401.

Evaluate:`int((e/x)^x+(x/e)^x)1nx dxdot`

Answer» Let `((x)/(e))^(x)=t`
or `(((x)/(e))^(x) "In"(x)/(e)+x*((x)/(e))^(x-1)*(1)/(e))dx=dt`
or `((x)/(e))^(x)("In"(x)/(e)+1)dx=dt`
or `((x)/(e))^(x) "In"xdx=dt`
So, ` I=int((1)/(t^(2))+1)dt= -(1)/(t)+t+C`
`=((x)/(e))^(x)-((e)/(x))^(x)+C`
Discussion
402.

Evaluate `int sin(logx)dx`.

Answer» Let `log x =t.` Then,
`x=e^(t)impliesdx=e^(t)dt`
` :. I =int sin(logx)dx`
`=int sin t e^(t) dt`
`=(e^(t))/(2)(sint-cost)+C`
Hence, `int sin(logx)dx=(x)/(2) [sin(logx)-cos(logx)]+C`
Discussion
403.

Evaluate `intsqrt(1+3x-x^2)dx`.

Answer» `intsqrt(1+3x-x^2)dx`
`=intsqrt(1+(9)/(4)-(x^(2)-2x((3)/(2))+(9)/(4)))dx`
`=int sqrt(((sqrt(13))/(2))^(2)-(x-(3)/(2))^(2))dx`
`=(x-(3)/(2))/(2) sqrt(1+3x-x^(2))+(13)/(4xx2)"sin"^(-1)((x-(3)/(2))/((sqrt(13))/(2)))+C`
`=(2x-3)/(4)sqrt(1+3x-x^(2))+(13)/(8)"sin"^(-1)((2x-3)/(sqrt(13)))+C`
Discussion
404.

Evaluate:`int(sinx)/(sin3x)dxdot`

Answer» `I=int (sinx)/(sin3x)dx=int(sinx)/(3sinx-4sin^(3)x)dx`
`=int(1)/(3-4sin^(2)x)dx`
`=int(sec^(2)x)/(3sec^(2)x-4tan^(2)x)dx`
` " " [ "Dividing " N^(r) and D^(r) " by " cos^(2)x]`
Putting `tanx=t and sec^(2)xdx=dt,` we get
`I=int(dt)/(3(1+t^(2))-4t^(2))=int(dt)/(3-t^(2))=int(1)/((sqrt(3))^(2)-t^(2))dt`
`=(1)/(2sqrt(3))log|(sqrt(3)+t)/(sqrt(3)-t)|+C=(1)/(2sqrt(3))log|(sqrt(3)+tanx)/(sqrt(3)-tanx)|+C`
Discussion
405.

Evaluate:`int1/(sqrt(3x+4)-sqrt(3x+1))dxdot`

Answer» `int(1)/(sqrt(3x+4)-sqrt(3x+1))dx`
`=int(sqrt((3x+4))+sqrt((3x+1)))/((sqrt((3x+4))+sqrt((3x+1)))(sqrt((3x+4))-sqrt((3x+1))))dx`
`=int(sqrt((3x+4))+sqrt((3x+1)))/((3x+4)-(3x+1))dx`
`=(1)/(3)sqrt(3x+4)+sqrt(3x+1)dx`
`=(1)/(3)sqrt(3x+4)dx+(1)/(3)sqrt(3x+1)dx`
`=(1)/(3){((3x+4)^(3//2))/(3xx(3)/(2))}+(1)/(3){((3x+1)^(3//2))/(3xx(3)/(2))}+C`
`=(2)/(27){(3x+4)^(3//2)+(3x+1)^(3//2)}+C`
Discussion
406.

Evaluate:`intsin^(-1)sqrt(x/(a+x))dxdot`

Answer» `I=int sin^(-1)sqrt((x)/(a+x))dx`
Let `x=a tan^(2) theta " or " dx=2a tan theta sec^(2)theta d theta`
`:. intsin^(-1)sqrt(sin^(2)theta)2a sec^(2)theta tan theta d theta`
`=2a int underset(I)(underbrace(theta)) underset(II)(underbrace(sec^(2) theta tan theta)) d theta`
`=2a[theta(tan^(2)theta)/(2)-int(tan^(2)theta)/(2) d theta]`
`=a[ theta tan^(2)theta-int(sec^(2) theta-1) d theta]`
`=a(theta tan^(2) theta-tan theta + theta]+c, " where " theta ="tan"^(-1)sqrt((x)/(a))`
Discussion
407.

The value of `inta^sqrt(x)/sqrt(x)` dx is equal toA. `a^sqrt(x)/sqrt(x)+C`B. `(2a^sqrt(x))/(lna)+C`C. `2a^sqrt(x).ln a+C`D. `2a^sqrt(x)+C`

Answer» Correct Answer - B
Discussion
408.

Evaluate `int(secx+tanx)^(2)dx`

Answer» Correct Answer - `2(secx+tanx)-x+C`
`int(secx+tanx)^(2)dx=int(sec^(2)x+tan^(2)x+2secxtanx)dx`
`=int(2sec^(2)x-1+2secxtanx)dx`
`=2(secx+tanx)-x+C`
Discussion
409.

Evaluate:`inta^(m x)b^(n x)dx`

Answer» Correct Answer - `(a^(m)b^(n))^(x)/(log(a^(m)b^(n)))+C`
`I=inta^(mx)b^(nx)dx`
`=int(a^(m)b^(n))^(x)dx`
`=(a^(m)b^(n))^(x)/(log(a^(m)b^(n)))+C`
Discussion
410.

If`int1/(x+x^5)dx=f(x)+c ,t h e ne v a l u a t eint(x^4)/(x+x^5)dxdot`

Answer» Correct Answer - `logx-f(x)+C`
`int(x^(4)dx)/(x+x^(5))=int((x^(4)+1)dx)/(x+x^(5))-intdx/(x+x^(5))`
`=int((x^(4)+1)dx)/(x(1+x^(4)))-intdx/(x+x^(5))`
`=intdx/x-intdx/(x+x^(5))`
`=logx-f(x)+C`
Discussion
411.

Evaluate:`int(secx)/(secx+tanx)dxdot`

Answer» `int(secx)/(sec x +tan x)dx=int(secx(secx-tanx))/((secx+tanx)(secx-tanx))dx`
`=int(sec^(2)x-secxtanx)/(sec^(2)x-tan^(2)x)dx`
`=int sec^(2)xdx-intsecx tanx dx`
`=tanx - secx +C`
Discussion
412.

Evaluate `int(tanx)/(secx+tanx)dx`

Answer» Correct Answer - `secx-tanx+x+C`
`int(tan)/(secx+tanx)dx=int(tanx(secx-tanx))/((secx+tanx)(secx-tanx))dx`
`=int(tanx(secx-tanx))/(sec^(2)x-tan^(2)x)dx`
`=int(secxtanx-tan^(2)x)dx`
`=int(secxtanxdx-int(sec^(2)x-1)dx`
`=int(secxtanxdx-intsec^(2)xdx+int1dx`
`=secx-tanx+x+C`
Discussion
413.

The integral `int(sec^2x)/((secx+tanx)^(9/2))dx`equals (for some arbitrary constant `K)dot``-1/((secx+tanx)^((11)/2)){1/(11)-1/7(secx+tanx)^2}+K``1/((secx+tanx)^(1/(11))){1/(11)-1/7(secx+tanx)^2}+K``-1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K``1/((secx+tanx)^((11)/2)){1/(11)+1/7(secx+tanx)^2}+K`A. `-(1)/((secx+tanx)^(11//2)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`B. `(1)/((secx+tanx)^(1//11)){(1)/(11)-(1)/(7)(secx+tanx)^(2)}+K`C. `-(1)/((secx+tanx)^(11//2)){(1)/(11)+(1)/(7)(secx+tanx)^(2)}+K`D. `(1)/((secx+tanx)^(11//2)){(1)/(11)+(1)/(7)(secx+tanx)^(2)}+K`

Answer» Correct Answer - C
`I=int(sec^(2)x)/((secx+tanx)^(9//2))dx`
Let ` sec x +tanx=t`
` or sec x -tan x=1//t`
Now, `(secx tanx+sec^(2)x)dx =dt`
`or secx(secx+tanx)dx=dt`
` or secx dx=(dt)/(t),(1)/(2)(t+(1)/(t))=sec x`
` :. I=(1)/(2)int ((t+(1)/(t)))/(t^(9//2))(dt)/(t)`
`=(1)/(2)int(t^(-9//2)+t^(-13//2))dt`
`=(1)/(2)[(t^(-9//2+1))/(-(9)/(2)+1)+(t^(-13//2+1))/(-(13)/(2)+1)]+K`
`=(1)/(2)[(t^(-7//2))/(-(7)/(2))+(t^(-11//2))/(-(11)/(2))]+K`
`= -(1)/(7) t^(-7//2)-(1)/(11)t^(-11//2)+K`
`= -(1)/(7) (1)/(t^(7//2))-(1)/(11)(1)/(t^(11//2))+K`
`= -(1)/(t^(11//2))((1)/(11)+(t^(2))/(7))+K`
`= -(1)/((secx+tan x)^(11//2)){(1)/(11)+(1)/(7)(sec x+tanx)^(2)}+K`
Discussion
414.

`int{1+2tanx(tanx+secx)}^(1//2)dx` is equal toA. `"ln"|secx(secx-tanx)|+C`B. `"ln"|"cosec "x(sec x+tanx)|+C`C. `"ln"|secx(secx+tanx)|+C`D. `-"ln"|cosx(secx-tanx)|+C`

Answer» Correct Answer - CD
Discussion