Integrating by substitutions q's (1 Viewer)

[Libbster]

Well just my luck I was away when we did this in class and I am stuck on a few questions. I was wondering if I could get some help! I can do 34 out of the 39 q's in the book but on these ones I have no idea how they got to the answer! This is from the 3 unit coroneos book

Integrate sin2x/ (2+ 3cos^2 x) dx let u = 2 + 3 cos^2 x
The answer is -1/3 ln(2 + cos^2 x)

Integrate 10^ tanx . sec^2 x dx (ie 10 to the power of tanx multiplied by sec squared x) let u = tan x
The answer is 10^tanx / ln10

Integrate x/ (x + 1) dx let x = u^2 -1
The answer is (x+1) - ln(x + 1)

Integrate x / (4 - x^2)^1/2 dx let x = 2 sin @
The answer is -(4 - x^2)^1/2

Integrate dx/ (4 - x^2)^3/2 let x= 2 sin@
The answer is 1/4x / (4 - x^2)^1/2

Sorry for so many q's but we have an assessment task on Tuesday !

 

[FinalFantasy]

Integrate sin2x/ (2+ 3cos^2 x) dx let u = 2 + 3 cos^2 x
The answer is -1/3 ln(2 + cos^2 x)

let u=2+3cos²x
du\dx=-3(2cosxsinx)=-6sinxcosx=-3sin2x
int. sin2x\(2+3cos²x) dx
=-1\3int. -3sin2x\(2+3cos²x) dx
=-1\3 int. du\u
=-1\3 ln u+C
=-1\3 ln(2 + 3cos^2 x) +C

 

[FinalFantasy]

Integrate 10^ tanx . sec^2 x dx (ie 10 to the power of tanx multiplied by sec squared x) let u = tan x
The answer is 10^tanx / ln10

I=int. 10^tanx sec²x dx
let u=tanx
du\dx=sec²x
.: I=int. 10^u du=(1\ln10)* 10^u+C

ntegrate x/ (x + 1) dx let x = u^2 -1
The answer is (x+1) - ln(x + 1)

let x=u²-1
dx=2u du
int. x\(x+1) dx =int. 2u(u²-1)\u² du=2int. (u-1\u) du=2(u²\2)-2ln u
=(x+1)-2ln (x+1)^1\2=x+1-ln(x+1)

 

[Trev]

Integrate x/ (x + 1) dx let x = u^2 -1
The answer is (x+1) - ln(x + 1)

int. x/(x+1).dx
x = u² - 1
dx/du = 2u
dx = 2udu

∴ int. x/(x+1).dx
= int. (u² - 1)/(u²).2u.du
= 2 int. (u - 1/u).du
= u² - 2lnu + C
= (x+1) - 2ln[(x+1)^1/2] + C
= (x+1) - ln(x+1) + C

 

[FinalFantasy]

Integrate x / (4 - x^2)^1/2 dx let x = 2 sin @
The answer is -(4 - x^2)^1/2

let x=2sin@
dx=2cos@d@
int. x\(4-x²)^1\2 dx=int. (2cos@)2sin@\(4-4sin²@)^1\2 d@
=4int. cos@sin@d@\2sqrt(cos²@)=2int. sin@ d@=-2cos@+C
x\2=sin@
sin²@=1-cos²@
cos²@=1-sin²@=1-x²\4
.: I=sqrt(-2(1-x²\4))=sqrt(-2+x²\4)

 

[FinalFantasy]

Integrate dx/ (4 - x^2)^3/2 let x= 2 sin@
The answer is 1/4x / (4 - x^2)^1/2

I=dx\(4-x²)^(3\2)
let x=2sin@
dx=2cos@ d@

I=int. 2cos@ d@\(4-4sin²@)^3\2
=int. 2cos@ d@\8(cos²@)^(3\2)
=1\4 int. 1\cos²@ d@=1\4int. sec² d@
=1\4 tan @+C
sin@=x\2
draw a diagram so da remaining side is sqrt(4-x²)
I=1\4 x\sqrt(4-x²)

 

[nick1048]

Integrate sin2x/ (2+ 3cos^2 x) dx let u = 2 + 3 cos^2 x
The answer is -1/3 ln(2 + cos^2 x)

let u=2+3cos²x
du\dx=-3(2cosxsinx)=-6sinxcosx=-3sin2x
int. sin2x\(2+3cos²x) dx
=-1\3int. -3sin2x\(2+3cos²x) dx
=-1\3 int. du\u
=-1\3 ln u+C
=-1\3 ln(2 + cos^2 x) +C

ok i get the same answer and working as this but the last line I get u = 2 + 3cos^2 x not cos^2 x... Why does that just go away??? I mean ur substituting u back in.

.'. answer is -1/3 ln (2 + 3cos^2x) + C

 

[FinalFantasy]

ok i get the same answer and working as this but the last line I get u = 2 + 3cos^2 x not cos^2 x... Why does that just go away??? I mean ur substituting u back in.

.'. answer is -1/3 ln (2 + 3cos^2x) + C

i just copied and paste or something, or maybe typed it wrong LoL
fixed it

 

[Libbster]

Thank you so much!!!!