How do you integrate #\int \frac { t ^ { 2} } { \sqrt { 1- t ^ { 6} } } d t#?

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Answer 1 · 2017-09-05T01:42:59

# int \ t^2/(sqrt(1-t^6)) \ dt = 1/3 \ arcsin(t^3) + C #

We seek:

# I = int \ t^2/(sqrt(1-t^6)) \ dt #

Let us perform a substitution;

put #u=t^3 => (du)/dt = 3t^2 #

Substituting into the integral we get:

# I = 1/3 \ int \ (3t^2)/(sqrt(1-t^6)) \ dt #
# \ \ = 1/3 \ int \ (1)/(sqrt(1-u^2)) \ du #

Which is a standard integral, and so:

# I = 1/3 \ arcsin(u) + C #

And, restoring the substitution, we get:

# I = 1/3 \ arcsin(t^3) + C #