Integrate: x(sqr.root: 1 - x^4)dx?

Question

Please show work using trig. substitution.
-thanks

Answer 1

Integrate x/√(1 - x^4) with respect to x.

∫[x/√(1 - x^4)] dx
Let
sinθ = x²
cosθ dθ = 2x dx
cosθ dθ/2 = x dx
θ = arcsin(x²)

= (1/2)∫[cosθ/√(1 - sin²θ)] dθ

= (1/2)∫[cosθ/√(cos²θ)] dθ

= (1/2)∫[cosθ/cosθ] dθ

= (1/2)∫dθ

= θ/2 + C

= (1/2)arcsin(x²) + C

Answer 2

Partial fractions a hundred and one: First have 2 arbitrary values to precise the numerators of the two fractions- A and B a million/n(n+a million) = A/n + B/n+a million multiply the function via the denominator of the unique fee (in this occasion, n(n+a million)) a million=A(n+a million) + Bn make individual areas of the hot function = 0 via determining on values of n. as an occasion, via letting n=0, we get rid of the term Bn from the equation as any huge style x 0 = 0 a million = A, so A =a million in addition, permit n =-a million so as that A(n+a million) = 0, leaving Bn=a million -B=a million ergo B=-a million for this reason, we evaluate to our unique function and notice that via substituting the values of A and B that we've discovered a million/n(n+a million) = A/n + B/n+a million turns into: a million/n(n+a million) = a million/n + -a million/n+a million) = a million/n - a million/n+a million then combine the hot partial fraction format of the expression, utilizing the guideline that a million/x integrates to ln(x) so a million/n -a million/n+a million integrates via n to ln(n) - ln(n+a million) + c or ln(n/n+a million) + c in case you like to simplify.