What is Integration by Parts?
A technique to integrate the product of two functions, based on the product rule for differentiation.
Define 'u' in Integration by Parts.
A selected part of the integrand that is differentiated.
Define 'dv' in Integration by Parts.
A selected part of the integrand that is integrated.
What is 'du' in Integration by Parts?
The derivative of 'u' with respect to the variable of integration.
What is 'v' in Integration by Parts?
The antiderivative (integral) of 'dv'.
Steps to solve $\int xe^x , dx$ using Integration by Parts?
1. Choose $u=x$, $dv=e^x dx$. 2. Find $du=dx$, $v=e^x$. 3. Apply formula: $xe^x - \int e^x dx$. 4. Evaluate: $xe^x - e^x + C$.
Steps to solve $\int \ln(x) , dx$ using Integration by Parts?
1. Rewrite as $\int 1 \cdot \ln(x) , dx$. 2. Choose $u=\ln(x)$, $dv=dx$. 3. Find $du=\frac{1}{x}dx$, $v=x$. 4. Apply formula: $x\ln(x) - \int x(\frac{1}{x}) dx$. 5. Evaluate: $x\ln(x) - x + C$.
Steps to solve $\int x^2 \cos(x) , dx$?
1. Choose $u=x^2$, $dv=\cos(x)dx$. 2. Find $du=2x dx$, $v=\sin(x)$. 3. Apply formula: $x^2\sin(x) - \int 2x\sin(x) dx$. 4. Apply IBP again to $\int 2x\sin(x) dx$. 5. Final result: $x^2 \sin(x) + 2x \cos(x) - 2\sin(x) + C$.
Steps to evaluate $\int e^x \cos x , dx$?
1. Apply IBP twice, keeping $e^x$ as part of 'u' or 'dv' consistently. 2. Isolate the original integral using algebraic manipulation. 3. Solve for the original integral.
What are the differences between u-substitution and Integration by Parts?
u-substitution: Reverses the chain rule, simplifies composite functions. | Integration by Parts: Reverses the product rule, simplifies products of functions.
