Summary
Highlights
The video begins by introducing the three fundamental formulas for integrating functions that result in inverse trigonometric functions: arcsin(U/A), (1/A)arctan(U/A), and (1/A)arcsecant(U/A), along with the constant C.
The first example demonstrates how to find the anti-derivative of dx / sqrt(16 - x^2). The solution involves identifying U=x and A=4, then substituting these into the arcsin formula to get arcsin(x/4) + C.
The second example calculates the anti-derivative of 3 / (25 + x^2) dx. Here, U=x and A=5. After moving the constant 3 to the front, the arctan formula is applied, leading to (3/5)arctan(x/5) + C.
This example tackles 8 / (x * sqrt(4x^2 - 1)) dx. The key is to recognize this as an arcsecant form. U=2x and A=1. The video shows how to substitute DX and X in terms of DU and U, simplify, and then apply the arcsecant formula to derive 8arcsecant(2x) + C.
The example x / sqrt(1 - x^4) dx is solved. By setting U=x^2 and A=1, and substituting DX with du/(2x), the 'x' terms cancel, simplifying the integral to a basic arcsin form, resulting in (1/2)arcsin(x^2) + C.
This problem, x / (x^4 + 36) dx, is another arctan type. U=x^2 and A=6. After similar substitutions and cancellations as previous examples, the solution is (1/12)arctan(x^2/6) + C.
For x^3 / (x^2 + 1) dx, long division is used first due to the higher degree of the numerator. This transforms the integral into two parts: x dx and -x / (x^2 + 1) dx. The second part is solved using U-substitution, leading to x^2/2 - (1/2)ln(x^2 + 1) + C.
The anti-derivative of e^(3x) / (9 + e^(6x)) is calculated. A=3 and U=e^(3x). Following the arctan formula and substitutions, the answer is (1/9)arctan(e^(3x)/3) + C.
This example integrates (x - 3) / (x^2 + 1) dx by splitting it into two integrals: x / (x^2 + 1) dx and -3 / (x^2 + 1) dx. The first part uses U-substitution for a natural log, and the second part uses the arctan formula, resulting in (1/2)ln(x^2 + 1) - 3arctan(x) + C.
The integral dx / (x^2 - 4x + 7) requires completing the square in the denominator. This transforms the expression into (x - 2)^2 + 3. Then, U=x-2 and A=sqrt(3), leading to (1/sqrt(3))arctan((x-2)/sqrt(3)) + C.
This problem, x dx / (x^4 + 2x^2 + 2), also involves completing the square, resulting in (x^2 + 1)^2 + 1. Setting U=x^2+1 and A=1, the integral simplifies to (1/2)arctan(x^2+1) + C after cancellations.
The integration of 2x / (x^2 + 6x + 13) dx is solved by completing the square to get (x + 3)^2 + 4. This problem requires solving for X in terms of U (U=x+3, so x=U-3) and then splitting the integral into two parts: one for natural log and one for arctan. The final answer is ln(x^2 + 6x + 13) - 3arctan((x+3)/2) + C.
This complex problem, x / sqrt(9 + 8x^2 - x^4) dx, requires completing the square inside the square root, factoring out a negative. This leads to sqrt(25 - (x^2 - 4)^2). Here, A=5 and U=x^2 - 4, leading to the arcsin formula. The solution is (1/2)arcsin((x^2 - 4)/5) + C.
The final example is a definite integral from 3 to 6 of dx / ( (x - 3)^2 + 25). This is an arctan form. The limits of integration are changed from x-values to U-values (0 to 3), and then the arctan formula is applied. The definite integral evaluates to (1/5)arctan(3/5), as arctan(0) is 0.