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Calc 2 problem that i need help with.?
Intergrate (x^3)/(sqrt(x^2+4)) dx from 0 to 2. i know that you need to use trig sub. using x=a*tan(Θ)
i did this question probably like 5 times and my answer was (13*sqrt(2)-5)/ 3
but the right answer on my sheet says: 8/3*(2-sqrt(2))
The indefinite integral is (8(sec(Θ))^3)/3-(sec(Θ))
2 Answers
- mohanrao dLv 79 years ago
Trignometric substitution is not necessary
∫ x^3 dx /√(x^2 + 4) from 0 to 2
= ∫ x^2 (x dx ) /√(x^2 + 4)
let x^2 + 4 = t^2 when x = 0, t = 2 and when x = 2, t = 2√2
x^2 = t^2 - 4
2x dx = 2t dt
x dx = t dt
now the integral changes to
∫ (t^2 - 4) t dt / t
= ∫ (t^2 - 4) dt from 2 to 2√2
= (1/3)t^3 - 4t from 2 to 2√2
= (1/3)(16√2 - 8) - 4(2√2 - 2)
= (1/3) [ 16√2 - 8 - 24√2 + 24 ]
= (1/3) [16 - 8√2 ]
= 8/3 (2 - √2)
- Man of SteelLv 69 years ago
x=2tanv
v=theta
dx=2sec^2vdv
8tan^3v/[sqrt(4tan^2v+4)] * 2sec^2v
8tan^3v/[2secv] * 2sec^2v
8tan^3v*2secv
16tan^3v*secv
16*INT[tan^3v*secv]
16*INT[tan^2vtanv*secvdv]
16*INT[(sec^2v-1)tanvsecvdv]
16*INT[sec^2vtanvsecv-tanvsecv dv]
break it up.
16*INT[sec^2vtanvsecv] - 16*INT[secvtanv dv]
t-sub for first integral.
t=secv
dt=secvtanvdv
16*INT[t^2dt] - 16*INT[secvtanv dv]
16*(1/3*t^3) - 16secv
16/3*(secv)^3 - 16secv
recall: x=2tanv
x/2=tanv
secv=1/2*sqrt(x^2+4)
16/3*[1/2*sqrt(x^2+4)]^3 - 16*[1/2*sqrt(x^2+4)]
8/3*[sqrt(x^2+4)]^3 - 8*[sqrt(x^2+4)]
plug in 2
then plug in 0.
