When evaluating ∫ 2x·cos(x²) dx, which substitution is most effective?
Au = cos(x²)
Bu = x²
Cu = 2x
Du = sin(x²)
Setting u = x² gives du = 2x dx, so the integral becomes ∫ cos(u) du = sin(u) + C = sin(x²) + C. The derivative of the inner function x² appears in the integrand, which is exactly the chain-rule structure u-substitution exploits.
Question 2 True / False
When applying u-substitution to a definite integral ∫[a to b] f(g(x))·g'(x) dx, you can substitute u = g(x) and still use the original x-limits a and b.
TTrue
FFalse
Answer: False
Once you substitute u = g(x), the variable of integration changes to u, so the limits must also change to u-values: the lower limit becomes g(a) and the upper becomes g(b). Using the original x-limits with u-expressions is a category error that produces an incorrect answer. Alternatively, you may convert the antiderivative back to x before applying the original limits.
Question 3 Short Answer
Why is u-substitution considered the integration counterpart of the chain rule?
Think about your answer, then reveal below.
Model answer: The chain rule states d/dx[f(g(x))] = f'(g(x))·g'(x). U-substitution reverses this: when an integrand has the form f'(g(x))·g'(x), substituting u = g(x) and du = g'(x) dx turns it into ∫f'(u) du = f(u) + C = f(g(x)) + C — exactly undoing the chain rule.
Every integration technique is the reverse of a differentiation rule. U-substitution undoes the chain rule just as the power rule for integration undoes the power rule for differentiation. Recognizing which differentiation rule produced the integrand tells you which technique to apply.