hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #1
Applying substitution x = u^3, u = x^(1/3) with dx = 3*u^2*du, du = 1/3/x^(2/3)*dx
Applying substitution u = u1-1 with du=du1
Reverting substitution using u1 = 1+u
Reverting substitution using u = x^(1/3)
This problem is complete
>
Int((tan(x))^3,x);
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #2
Reduce the power on tan(x) using the identity tan(x)^2 = sec(x)^2-1.
Since the derivative of sec(x) is sec(x)*tan(x), we can use the substitution u = sec(x).
Applying substitution x = arcsec(u), u = sec(x) with dx = 1/u^2/(1-1/u^2)^(1/2)*du, du = sec(x)*tan(x)*dx
Reverting substitution using u = sec(x)
Since the derivative of cos(x) is -sin(x), we can use the substitution u = cos(x).
Applying substitution x = arccos(u), u = cos(x) with dx = -1/(1-u^2)^(1/2)*du, du = -sin(x)*dx
Reverting substitution using u = cos(x)
This problem is complete
>
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #3
Applying substitution x = -1/2*u, u = -2*x with dx = -1/2*du, du = -2*dx
This is one of the basic integrals.
Reverting substitution using u = -2*x
This problem is complete
>
Int(x^2 * ln(x), x);
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #4
Integrals of the form Int(a(x)*g(b(x))^k, x), where g is a logarithmic function, inverse trigonometric function, or inverse hyperbolic function, can often be simplified using integration by parts with u = g(b(x))^k.
This problem is complete
>
Int(sin(x)^3 *cos(x)^4, x);
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #5
Split off an even power of sin(x) and use the identity sin(x)^2 = 1-cos(x)^2.
Since the derivative of cos(x) is -sin(x), we can use the substitution u = cos(x).
Applying substitution x = arccos(u), u = cos(x) with dx = -1/(1-u^2)^(1/2)*du, du = -sin(x)*dx
Reverting substitution using u = cos(x)
Since the derivative of cos(x) is -sin(x), we can use the substitution u = cos(x).
Applying substitution x = arccos(u), u = cos(x) with dx = -1/(1-u^2)^(1/2)*du, du = -sin(x)*dx
Reverting substitution using u = cos(x)
This problem is complete
>
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #6
Hints:
1. Integrals involving expressions of the form (x^2+a^2)^n or sqrt(x^2+a^2)^n can often be simplified using the substitution x = a*tan(u).
2. Use the substitution 4+x^2 = u^2.
3. Integrals involving expressions of the form sqrt(x^2+a^2) can often be simplified using the substitution u = sqrt(x^2+a^2)-x.
Applying substitution x = 2*tan(u), u = arctan(1/2*x) with dx = (2+2*tan(u)^2)*du, du = 1/2/(1/4*x^2+1)*dx
Split off an even power of tan(u) and use the identity tan(u)^2 = sec(u)^2-1.
Since the derivative of sec(u) is sec(u)*tan(u), we can use the substitution u1 = sec(u).
Applying substitution u = arcsec(u1), u1 = sec(u) with du = 1/u1^2/(1-1/u1^2)^(1/2)*du1, du1 = sec(u)*tan(u)*du
Reverting substitution using u1 = sec(u)
Since the derivative of sec(u) is sec(u)*tan(u), we can use the substitution u1 = sec(u).
Applying substitution u = arcsec(u1), u1 = sec(u) with du = 1/u1^2/(1-1/u1^2)^(1/2)*du1, du1 = sec(u)*tan(u)*du
Reverting substitution using u1 = sec(u)
Reverting substitution using u = arctan(1/2*x)
This problem is complete
>
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #7
Hints:
1. Integrals involving expressions of the form (x^2+a^2)^n or sqrt(x^2+a^2)^n can often be simplified using the substitution x = a*tan(u).
2. Use the substitution 4+x^2 = u^2.
3. Integrals involving expressions of the form sqrt(x^2+a^2) can often be simplified using the substitution u = sqrt(x^2+a^2)-x.
Applying substitution x = 2*tan(u), u = arctan(1/2*x) with dx = (2+2*tan(u)^2)*du, du = 1/2/(1/4*x^2+1)*dx
Make change of variable.
Applying substitution u = arccos(u1), u1 = cos(u) with du = -1/(1-u1^2)^(1/2)*du1, du1 = -sin(u)*du
Reverting substitution using u1 = cos(u)
Reverting substitution using u = arctan(1/2*x)
This problem is complete
>
>
hint := Hint(%);
Rule[hint](GetProblem());
while hint <>[] do
hint := Hint(GetProblem());
Rule[hint](GetProblem());
end do;
Creating problem #8
Applying substitution x = 1+u with dx=du
Reverting substitution using u = x-1
Applying substitution x = 1+u with dx=du
Reverting substitution using u = x-1
Rewrite the numerator in a form which contains the derivative of the denominator
Note that the derivative of 4+x^2 is 2*x, so we can make a change of variable.
Applying substitution x = (u-4)^(1/2), u = 4+x^2 with dx = 1/2/(u-4)^(1/2)*du, du = 2*x*dx
Reverting substitution using u = 4+x^2
Integrals involving expressions of the form (x^2+a^2)^n or sqrt(x^2+a^2)^n can often be simplified using the substitution x = a*tan(u).
Applying substitution x = 2*tan(u), u = arctan(1/2*x) with dx = (2+2*tan(u)^2)*du, du = 1/2/(1/4*x^2+1)*dx