Having calculus problem evaluate the integral from ln2 to ln6 of 6e^tcosh t dt I came up with 60 + 3 ln 12?

Question

I am not confident in this result any suggestion welcome

Answer 1

∫ 6 e^t cosh t dt limit from log2 to log 6

2 cosh t = e^t + e^-t
3 ∫ e^t 2cosh t dt
3 ∫ e^t [e^t + e^-t] dt
3 ∫ [e^2t + e^0t] dt
3 ∫ e^2t dt + 3 ∫1 dt
(3/2) [e^2t] + 3 [t] apply limit

(3/2) [e^2{log6} - e^2{log2}] + 3 [log6 – log2]
(3/2) [e^{log(6)^2}- e^{log(2)^2}] + 3 [log (6/2)] e^log =1
(3/2) [(6)^2}- (2)^2}] + 3 [log (3)]
(3/2) [32] + 3 [log (3)]
= 48 + 3 [log (3)] ans
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what you did is that you applied limits with + sign in [log6 + log2] = log 12
and similarly [(6)^2}+ (2)^2}] =40 then 40*3/2 =60

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Your 2nd integral

In this integral, the first answerer is right, but solution is much shorter
∫ ( -e^(cosˉ¹(x)) dx / √(1 - x^2) )

let x = cos t or t = cos^-1 x
so dx = - sin t dt

∫ ( -e^(t) [- sin t dt / sin t
∫ e^(t) dt
[e^(t)]
[e^( cos^-1 x)] answer
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Your last integral: there are conflicts as to what is right answer/ let me use new method

∫ e^(-x) cos 3x dx limit from 0 to infinity

cos 3x = [e^(3ix) + e^(-3ix)] / 2 and
sin 3x = [e^(3ix) - e^(-3ix)] / (2i)

∫ e^(-x) [e^(3ix) + e^(-3ix)] / 2 dx
(1/2) ∫ [e^(3ix - x) + e^-(3ix + x)] dx integrate

(1/2) {[e^(3ix - x)] / (3i - 1)} – (1.2){[e^ -(3ix + x)] / (3i + 1)}
(1/2) e^(- x) {[e^(3ix)] / (3i - 1)} – (1.2){[e^ -(3ix)] / (3i + 1)}

(1/2) e^(- x) {[e^(3ix)] (3i + 1) - (3i - 1)[e^(-3ix)]} / (-9 - 1)}
- (1/20) e^(- x) {[e^(3ix)] (3i + 1) - (3i - 1)[e^(-3ix)]}
- (1/20) e^(- x) {3i [e^(3ix) - e^(-3ix)] +1 [e^(3ix) - e^(-3ix)] }

- (1/20) e^(- x) {3i * 2i sin (3x) + 2 cos (3x) }
- (1/20) e^(- x) { -6 sin (3x) + 2 cos (3x) }

(1/10) e^(- x) {3 sin (3x) - cos (3x)}

as x>> infinity {3 sin (3x) - cos (3x)}/ e^(x) >>> 0

so lower limit is left x>> 0 it will begin with – sign

- {(1/10) e^(- 0) {3 sin (0) - cos (0)}
- {(1/10) 1{0 - 1}
= 1/10
There is no restriction of complex - i – form if one can collect back all real and imaginary parts together.

Answer 2

Been a at the same time as; combine via areas enable u=x^2-a million then du= (2x)dx remedy int (u^4)du from 0 to a million provides (u^5)/5 then substititute returned the u: that's [(x^2-a million)^5]/5 evaluated from 0 to a million: whilst x=a million the int is 0, whilst x=0 the int is -a million/5 -a million/5 - 0 = -a million/5 answer

Answer 3

I assume that you started by replacing cosht by its exponential form (e^t + e^-t)/2. This gives integral of 3e^(2t) +3. I get a final answer of 48 + 3ln3 but I must admit to doing it very quickly so this could be wrong.