What types of questions are covered in Exercise 3.2?

Types One, Two, Three, and Four questions.

What is the formula for Type Two substitution?

x upon f a d one results in log mode denominator plus c.

Dividing the numerator and denominator by e^x.

It aids in solving mathematical equations involving trigonometric functions.

By integrating specific expressions and solving for the coefficients.

Substitution Type Two in integration involves replacing elements with a soft corner and focusing on derivatives, leading to log mode denominator solutions.
The process of solving integration questions may require double substitutions, dividing by e^x for simplification, and adjusting angles in numerators and denominators, culminating in complex expressions that involve logarithmic functions and trigonometric identities.

What types of questions are covered in Exercise 3.2?
Types One, Two, Three, and Four questions.
What is the formula for Type Two substitution?
x upon f a d one results in log mode denominator plus c.
How can one simplify the integration of e^3x?
Dividing the numerator and denominator by e^x.
What is the significance of adjusting angles in numerators and denominators?
It aids in solving mathematical equations involving trigonometric functions.
How are A and B values found in equations?
By integrating specific expressions and solving for the coefficients.

00:00

The video covers Exercise 3.2 A after discussing basic complete integration.
Type One Basic questions were covered in the previous exercise.
Type Two, Three, and Four questions will be addressed in this exercise.
Substitution Type Two involves replacing troublesome elements with a soft corner for solving integration questions.
The formula x upon f a d one results in log mode denominator plus c.
The derivative of the substitution or denominator is crucial in Type Two questions.
The derivative of log x is 1 up x, leading to log mode denominator plus c.
Double substitution may be necessary for certain questions.
Integrating e3x results in log mode denominator plus c.
Dividing the numerator and denominator by e^x simplifies the integration process.

30:35

The text discusses mathematical equations involving denominators and numerators.
It mentions separating a whole square dinominette and the resulting cut.
The whole square of -2x / x s + 1 is dx1 + x s.
The solution involves putting what's inside brackets, either in seconds or first.
The process includes 1 upon x sp 1dx minus integration 2x upon x sk pw.
Derivatives and integrals are applied to solve the equations.
The text delves into the integration of various expressions.
It explains the process of adjusting angles in the numeration and denominators.
The method of finding A and B values in equations is detailed.
The final answer involves integrating specific expressions to derive the solution.

01:03:10

Numeration is equation number one, with its value placed in the numeration.
Integration of the equation results in a complex expression involving x, c, and derivatives.
Solving for the value of a in the equation leads to a being determined as 5.
Substituting the values of a and b back into the equation results in a final answer involving logarithmic functions.
The process involves intricate mathematical manipulations and the application of various trigonometric identities to arrive at the final solution.