Question 211741
a) &#8747;( (x^2x) / ((3 + x^3)^2) )  Substitution, u = x^3<br>
I assume this is supposed to be:
a) &#8747; {{{( ((x^2)dx) / ((3 + x^3)^2) )}}}  Substitution, u = x^3
Actually a better substitution would be: {{{u = 3 + x^3}}} but what you were given will still work. We start by finding the derivative of both sides of {{{u = x^3}}}:
{{{(du)/(dx) = 3x^2}}}
Multiplying both sides of this by dx we get:
{{{du = 3x^2*dx}}}
Looking at the original integral we can see the {{{x^2*dx)}}} but not the 3. So we can "move" the 3 by multiplying both sides by {{{1/3}}} (or divide both sides by 3):
{{{(1/3)du = x^2*dx}}}
Now we can substitute u for {{{x^3}}} and {{{(1/3)du}}} for {{{x^2*dx}}} giving:
a) &#8747; {{{( ((1/3)du) / ((3 + u)^2) )}}}
This can be rewritten as:
a) &#8747; {{{(1/3)((3 + u)^-2)du)}}}
If the integral above is not obvious to you, then use a second substitution: v = 3+u. Either way we should get:
{{{(1/3)(1/(-2 + 1))(3 + u)^(-2 + 1) + C = ((-1)/3)(3 + u)^-1 + C = (-1)/(3(3 + u)) + C = (-1)/(9 + 3x^3) + C}}}<br>
b) &#8747; \( x^7 cos x^8dx)  Substitution  u = x^8
I assume the "\" is a typo and that the integral is:
b) &#8747; {{{(x^7*cos(x^8)*dx)}}}
Again we start by finding the derivative of {{{u = x^8}}}:
{{{(du)/(dx) = x^7}}}
Multiplying both sides by dx:
{{{du = x^7*dx}}}
Using the Commutative Property on the original integral we can see the right side of the above equation in the integral:
&#8747; {{{(cos(x^8) * x^7*dx)}}}
So we can now substitute u for {{{x^8}}} and du for {{{x^7*dx}}}:
&#8747; {{{(cos(u) * du)}}}
This is a pretty simple integral to find:
{{{sin(u) + C = sin(x^8) + C}}}