Integration by U-Substitution

In the video, the speaker explains how to integrate using u-substitution, focusing on definite integrals.

To integrate 4x(x² + 5)³, set u = x² + 5, and du = 2x dx, then substitute u and du into the original equation, cancel out the x terms, and solve for the antiderivative.

The antiderivative is ½u⁴ + C, and replacing u with x² + 5 gives the final answer.

To integrate 8cos(4x) dx, set u = 4x, du = 4 dx, substitute u and du into the original equation, and solve for the antiderivative.

The antiderivative is 2sin(u) + C, and replacing u with 4x gives the final answer.

The key to u-substitution is identifying what u and du are, and then following the process.

To integrate x³e^(x⁴), set u = x⁴, du = 4x³ dx, substitute u and du into the original equation, and solve for the antiderivative.

The antiderivative is ¼e^u + C, and replacing u with x⁴ gives the final answer.

To integrate 8x√(40 - 2x²) dx, set u = 40 - 2x², du = -4x dx, substitute u and du into the original equation, and solve for the antiderivative.

The antiderivative is -⁴/₃u³² + C, and replacing u with 40 - 2x² gives the final answer.

To integrate x³/(2 + x⁴)², set u = 2 + x⁴, du = 4x³ dx, substitute u and du into the original equation, and solve for the antiderivative.

The antiderivative is -¹/₄u⁻¹ + C, and replacing u with 2 + x⁴ gives the final answer.

To integrate sin⁴(x)cos(x) dx, set u = sin(x), du = cos(x) dx, substitute u and du into the original equation, and solve for the antiderivative.

The antiderivative is ¹/₅u⁵ + C, and replacing u with sin(x) gives the final answer.

To integrate √(5x + 4), set u = 5x + 4, du = 5 dx, and substitute u and du into the original equation, but no further simplification is possible.- The problem involves substituting u for 5x + 4 and dx for du/5 to solve the integral.

The substitution is used to integrate the square root of u with respect to x, which becomes u^(1/2) du/5.

Applying the power rule, the antiderivative is found to be (2/15)u^(3/2) + C.

The final answer is obtained by replacing u with 5x + 4, resulting in (2/15)(5x + 4)^(3/2) + C.

In the next problem, u is substituted for 3x + 2, and dx is substituted for du/3.

The goal is to eliminate the x variable, so x is isolated by solving for x in the expression u - 2 = 3x, resulting in x = (1/3)u - (2/3).

The substitutions are then used to integrate the expression, which involves rewriting the square root of u as u^(1/2) and dx as du/3.

The antiderivative is found by distributing u^(1/2) to u and u - 2, and then integrating each term separately.

The final answer is obtained by replacing u with 3x + 2, resulting in (2/45)(3x + 2)^(5/2) - (4/27)(3x + 2)^(3/2) + C.

In the third problem, u is substituted for 4x - 5, and dx is substituted for du/4.

The goal is to eliminate the x variable, so x is isolated by solving for x in the expression u + 5 = 4x, resulting in x = (1/4)u + (5/4).

The substitutions are then used to integrate the expression, which involves rewriting the square root of u as u^(1/2) and dx as du/4.

The antiderivative is found by distributing u^(1/2) to u and u + 5, and then integrating each term separately.

The final answer is obtained by replacing u with 4x - 5, resulting in (1/20)(4x - 5)^(5/2) + (5/12)(4x - 5)^(3/2) + C.