Reverse the chain rule: substitute u = inside function, then du absorbs the inside's derivative.
Walk through
Step 1 of 5
Spot the inside function
In ∫ 2x · cos(x²) dx, the awkward part is cos(**x²**). Set u = x² — the function tucked inside another.
Substitution is the chain rule run backwards. If you ever see f(g(x))·g′(x), substitution will collapse it.
For definite integrals — change the limits
When you switch to u, also convert the bounds: in ∫₀^(√π) 2x cos(x²) dx with u = x², the limits become u = 0 and u = π, giving ∫₀^π cos u du directly. No need to switch back to x.
Your turn
∫ 3x² · (x³ + 1)⁵ dx — what is u?
Try it
∫ 2x · cos(x²) dx
Let u = x². du = 2x dx. ∫ cos u du = sin u + C = sin(x²) + C.