Question

Solve (5 questions)?

(Only use partial fractions where possible)

1. `\intx\sinx\cosxdx`
2. `\int(x^P)/(x(x^(2P)+1)dx`
3. `\int(\lnx)/(x\sqrt(1+(\lnx)^2)` `dx`
4. `\int\sqrt(1+e^x)dx`
5. `\int(x+\sin^-1(x))/(\sqrt(1-x^2))` `dx`

Answer 1

`int xsinxcosxdx = (sin2x-2xcos2x)/8 +C`

`int x^P/(x(x^(2P)+1)) dx = 1/P arctanx^P +C`

`int lnx/(xsqrt(1+(lnx)^2))dx = sqrt(1+(lnx)^2)+C`

`int sqrt(1+e^x)dx = 2sqrt(1+e^x) +ln ((sqrt(1+e^x)-1)/(sqrt(1+e^x)+1))+C`

`int (x+arcsinx)/sqrt(1-x^2)dx= 1/2arcsin^2x - sqrt(1-x^2)+C`

Explanation:

Not sure we really need partial fractions except in one case:

`(1)`

`int xsinxcosxdx = 1/2 int xsin2x dx`

integrate by parts:

`int xsinxcosxdx = -1/4 int xd (cos2x)`

`int xsinxcosxdx = -(xcos2x)/4 + 1/4 int cos2xdx`

`int xsinxcosxdx = -(xcos2x)/4 + 1/8 sin2x +C`

`int xsinxcosxdx = (sin2x-2xcos2x)/8 +C`

`(2)`

`int x^P/(x(x^(2P)+1)) dx = int x^(P-1)/(x^(2P)+1) dx`

Substitute `t=x^P`, `dt = Px^(P-1)`

`int x^P/(x(x^(2P)+1)) dx = 1/P int dt/(t^2+1) = 1/P arctant+C`

`int x^P/(x(x^(2P)+1)) dx = 1/P arctanx^P +C`

`(3)`

`int lnx/(xsqrt(1+(lnx)^2))dx`

Substitute `t = 1+(lnx)^2`, `dt =2lnx/x dx`

`int lnx/(xsqrt(1+(lnx)^2))dx = int dt/(2sqrt(t)) =sqrtt + C`

`int lnx/(xsqrt(1+(lnx)^2))dx = sqrt(1+(lnx)^2)+C`

`(4)`

`int sqrt(1+e^x)dx`

Substitute `t=sqrt(1+e^x)`, `dt= e^x/(2sqrt(1+e^x))dx =(t^2-1)/(2t)dx`

`int sqrt(1+e^x)dx = 2int t^2/(t^2-1)dt`

`int sqrt(1+e^x)dx = 2int (t^2-1+1)/(t^2-1)dt`

`int sqrt(1+e^x)dx = 2int dt + 2int dt/(t^2-1)`

`int sqrt(1+e^x)dx = 2t + 2int dt/((t-1)(t+1))`

`2/((t-1)(t+1)) = A/(t-1)+B/(t+1)`

`2= A(t+1)+B(t-1) = (A+B)t+(A-B)`

`{(A+B = 0),(A-B=2):}`

`{(A=1),(B=-1):}`

`int sqrt(1+e^x)dx = 2t + int dt/(t-1)- int dt/(t+1)`

`int sqrt(1+e^x)dx = 2t + ln abs (t-1)- ln abs (t+1) +C`

`int sqrt(1+e^x)dx = 2sqrt(1+e^x) +ln ((sqrt(1+e^x)-1)/(sqrt(1+e^x)+1))+C`

`(5)`

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

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

`int x/sqrt(1-x^2)dx = - int (d(1-x^2))/(2sqrt(1-x^2)) = -sqrt(1-x^2)+C`

`int arcsinx/sqrt(1-x^2)dx = int arcsinx d(arcsinx) = 1/2arcsin^2x+C`

`int (x+arcsinx)/sqrt(1-x^2)dx= 1/2arcsin^2x - sqrt(1-x^2)+C`

Answer 2

`int xsinxcosx dx = 1/8sin(2x) - 1/4xcos(2x) + C`

Explanation:

Here is the answer to `1`. I would rewrite and then use integration by parts. Recall that `sin(2x) = 2sinxcosx`.

`I = int x(1/2sin(2x))dx`

`I = int (xsin(2x))/2dx`

`I = 1/2int xsin(2x) dx`

Now let `u= x` and `dv = sin(2x)dx`. We can immediately see that `du = dx`. Integrating `dv` is a little harder. Letting `n = 2x`, we get `dn = 2dx` and `dx= (dn)/2`. Therefore `v = -1/2cosn = -1/2cos(2x)`

Now recall that integration by parts is

`int udv =uv - int vdu`

`int xsin(2x)dx = -1/2xcos(2x) - int -1/2cos(2x)dx`

`int xsin(2x)dx= -1/2xcos(2x) + int 1/2cos(2x)dx`

`int xsin(2x)dx= -1/2xcos(2x) + 1/4sin(2x) + C`

`1/2int xsin(2x)dx= 1/8sin(2x) - 1/4xcos(2x) + C`

Hopefully this helps!