a. Integral (x^2)/(sqrt(1+(x^2)))
Would I separate these two into 2 separate integrals? Like: Integral of x^2 and the other integral of 1/sqrt(1+(x^2))
b. Integral (x^7)/(ln(x^4))dx
Do I use integration by parts for this?
I put u= lnx
du = 1/x
dv = x^7
v = (x^8)/8
It doesn't seem to work though.
Also, how would I do the first problem without sinhx?
(a) would you separate 6^2/√(6^2+1) like that? I think not.
use trig substitutions. Let
x = tanθ
1+x^2 = sec^2θ
dx = sec^2θ dθ
and you have
∫tan^2θ/secθ (sec^2θ dθ)
= ∫tan^2θ secθ dθ
= ∫secθ secθtanθ dθ
Now let u = secθ and you have
∫u du
(b) let
u = x^4
du = 4x^3 dx
and you have
∫x^4/ln(x^4) x^3 dx
= 1/4 ∫u/lnu du
That's not possible using elementary functions. I suspect a typo.
For part a, how do you know to convert that to tan and sec? Is there some formula?
Also, I got as my answer: 1/2sec^2(theta)
But the answer given is:
[1/2(x)(sqrt((x^2)+1))] - [1/2ln(x+(sqrt(1+(x^2))))] + C
you pick your substitutions because of identities
sin^2 + cos^2 = 1
cosh^2 = sinh^2 + 1
by dividing by sin^2 or cos^2 you can get the corresponding ones involving tan,sec,cot,csc, etc.
as for the answer, recall that the integral of sec is log|sec+tan|, and that the inverse hyperbolic functions also can be written as strange logs.
I'm glad you have such faith in my math, but I make mistakes like anyone else. Starting with
∫tan^2θ secθ dθ
I should have gone on to say
∫tan^2θ secθ dθ
∫sec^3θ - secθ dθ
Now, we know that secθ gives us log|secθ+tanθ| = log|x+√(1+x^2)|
For sec^3θ we use integration by parts
u = secθ
dv = sec^2θ dθ
du = secθtanθ dθ
v = tanθ, so we have
tanθsecθ - ∫secθtan^2θ dθ
now we're back to where we started, so we do it again and we wind up with your answer.
a. To evaluate the integral of (x^2)/(sqrt(1+(x^2))), you can use a technique called u-substitution. Let's set u = 1 + (x^2). Then, differentiate both sides with respect to x to find du/dx = 2x. Rearranging, we have du = 2x dx.
Now, substitute these values into the integral. The integral becomes ∫(x^2)/(sqrt(1+(x^2))) dx = ∫(1/u) du.
Next, simplify the integral: ∫(1/u) du = ∫u^(-1) du.
Applying the power rule for integration, we have ∫u^(-1) du = ln|u| + C.
Finally, substitute back u = 1 + (x^2) into the result: ln|1+(x^2)| + C. So the integral of (x^2)/(sqrt(1+(x^2))) is ln|1+(x^2)| + C, where C is the constant of integration.
b. To evaluate the integral of (x^7)/(ln(x^4)) dx, you can indeed use integration by parts. Recall the integration by parts formula:
∫u dv = uv - ∫v du
Let's choose u = ln(x^4) and dv = x^7 dx.
Differentiate u to find du/dx: du/dx = (1/x) * 4x^3 = 4x^2.
Integrate dv to find v: ∫x^7 dx = (x^8)/8.
Now apply the integration by parts formula:
∫(x^7)/(ln(x^4)) dx = (ln(x^4) * (x^8)/8) - ∫((x^8)/8 * 4x^2) dx.
Simplify the integral:
= (x^8 ln(x^4))/8 - (1/2) ∫x^10 dx.
Now integrate the remaining term: ∫x^10 dx = (x^11)/11.
Substituting everything back into the original equation:
= (x^8 ln(x^4))/8 - (1/2)((x^11)/11) + C,
where C is the constant of integration.