integration (1 Viewer)

[want2beSMART]

still having a bit of trouble with integration... =(

(a) Use integration by parts to find the integral of xe^(3x)dx

(b) Evaluate the integral between pi/4 and 0 (sinx)/(cos^3x)dx

(c) by completing the square, find the integral of dx/sqrt(5+4x-x^2)

(d) Use the substitution of x=2sin[theta] to find the integral from 1 to 0 x^2/sqrt(4-x^2)dx

 

[FinalFantasy]

still having a bit of trouble with integration... =(

(a) Use integration by parts to find the integral of xe^(3x)dx

(b) Evaluate the integral between pi/4 and 0 (sinx)/(cos^3x)dx

(c) by completing the square, find the integral of dx/sqrt(5+4x-x^2)

(d) Use the substitution of x=2sin[theta] to find the integral from 1 to 0 x^2/sqrt(4-x^2)dx

a)int. xe^(3x)dx
let u=e^3x and dv\dx=x
du\dx=3e^3x and v=x²\2
int. xe^(3x)dx=(1\2)x²e^(3x)-3\2int. x²e^(3x) dx
int. xe^(3x)dx(1+3\2)=(1\2)x²e^(3x)
int. xe^(3x)dx=(1\5)x²e^(3x)+C

b)pi/4 and 0 int. (sinx)/(cos^3x)dx
let u=cosx, du=-sinx dx

int. (sinx)/(cos^3x)dx=-int. du\u³
............

c)integral of dx/sqrt(5+4x-x^2)
int. dx\sqrt(-(x²-4x-5))=int. dx\sqrt(-((x-2)²-9)=int. dx\sqrt(9-(x-2)²)
=sin^-1 (x-2)\3+C

d)1 to 0 x^2/sqrt(4-x^2)dx
let x=2sin@
dx=2cos@ d@
int. 4sin²@2cos@d@\2sqrt(cos²@)
=int. 4sin²@ d@
=4\2int. (1-cos2@) d@
=1\2 (@-1\2sin2@)+C
den apply terminals

 

[JamiL]

b is the only 1 i can remember doin b4 n not bothered attampting ne others. 2 way u caan get it by parts or subsitution.
I=intergral sinx/(cosx)^3dx
let u=cos x
du=-sinxdx
ie I=-intergral du/(u^3)
= -intergral u^-3
= 1/(2u)
= 1/(2cosx) +c
or u can use parts
u=tan x ie du=(secx)^2dx
dv=(secx)^2dx ie v=tan x
I=intergral tanx(secx)^2dx (all i did was re arange I if u cant see it u shouldn be doin 4u... lol
ie (using parts) I= (tanx)^2 - intergral tanx(secx)^2dx
ie I=(tanx)^2 - I
ie 2I=(tanx)^2
I=[(tanx)^2]/2 + C
u chose which 1 u want... i did them as indefinant, but they just reqire subsitution in which u should get a half as ur answer... correct me if rong i did that in my head

 

[KFunk]

a)int. xe^(3x)dx
let u=e^3x and dv\dx=x
du\dx=3e^3x and v=x²\2
int. xe^(3x)dx=(1\2)x²e^(3x)-3\2int. x²e^(3x) dx
int. xe^(3x)dx(1+3\2)=(1\2)x²e^(3x)
int. xe^(3x)dx=(1\5)x²e^(3x)+C

I think you probably should have done it the other way round. How did you combine the (3\2∫ x²e^3x dx) term with the (∫xe^3xdx) term?

Anyhow, using the other way
∫ xe^3xdx
let u=x and dv/dx = e^3x
du/dx=1 and v = 1/3.e^3x

int. = 1/3.xe^3x - ∫ 1/3.e^3x
= 1/3.xe^3x - 1/9.e^3x +C

[I could be wrong but I suspect that this is the way to go about it]

 

[FinalFantasy]

ooo..
LoL
i thought it was an x term, not an x² term

 

[KFunk]

haha, yeah that's what I figured.