Indefinite integral are written in the form of int `f(x) dx = F(x) +C`
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
In the given problem: `int t/sqrt(1-t^4)dt` , we follow:` int f(t)dt =F(t) +C.`
The problem can be rewritten as:
`int (t *dt)/sqrt(1^2-(t^2)^2)`
This resembles the basic integration formula for inverse sine function:
`int (du)/sqrt(a^2-u^2) = arcsin(u/a) +C`
Using u-substitution, we let `u = t^2` then `du = 2t*dt or (du)/2= t*dt` .
Note: `a^2 = 1` then` a = 1`
The indefinite integral will be:
`int (t *dt)/sqrt(1^2-(t^2)^2)=int ((du)/2)/sqrt(1^2-(u)^2)`
Applying the basic property of integration: `int c*f(x)dx = c int f(x) dx` , we get:
`(1/2) int (du)/sqrt(1^2-u^2)`
Applying the basic integral formula for inverse sine function:
`(1/2) int (du)/sqrt(1^2-u^2)=1/2arcsin(u/1) +C`
` =1/2arcsin(u)+C`
Plug-in ` u=t^2 ` in `1/2arcsin(u) +C` to express the indefinite intergral in terms of `int f(t)dt=F(t)+C` :
`int t/sqrt(1-t^4)dt =1/2arcsin(t^2) +C`
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