Integration by Substitution

For a direct reverse-chain-rule substitution, choose an inner expression g(x) whose derivative occurs beside it, perhaps multiplied by a constant. Set u = g(x) and du = g'(x) dx. To find that derivative, use the plain derivatives: the derivative of x to a power n is n times x to the power n minus 1, the derivative of ln x is 1 over x, and the derivative of sin x is cos x. For 3x²sin(x³) dx, u = x³ changes the integrand to sin(u) du. State the transformed expression you want to obtain. More than one substitution may work.

Compute the differential by differentiating the full expression for u: du = u'(x) dx. For u = 5x²+1, du = 10x dx. A constant added inside u contributes zero to its derivative. For a linear substitution u = ax+b with a nonzero, dx = du/a.

Convert each bound with the actual substitution. For u = x²+5, x from 2 to 3 gives u from 9 to 14. A decreasing substitution can reverse the bound order; keep that order with its differential sign. Alternatively, back-substitute the antiderivative and use the original x bounds. Either route is valid; do not mix variables.

Without a matching derivative factor, the direct shortcut may leave an extra factor. On a branch where x > 0, u = x² gives x = sqrt(u) and dx = du/(2sqrt(u)). Thus changing variables is still possible, but it need not produce a basic integral. Missing a factor alone does not prove that no elementary antiderivative exists.

Check the entire transformed differential, not just the inner expression. On x > 0, u = x⁴ in e^(x⁴) dx leaves e^u du/(4u^(3/4)). The extra factor prevents reduction to integral e^u du by this direct shortcut. To check a proposed antiderivative, differentiate it and compare the full result with the integrand.

Using radians, u = sin x gives du = cos x dx. Thus a product of a power of sin x and cos x dx becomes the corresponding power of u. For a definite integral, evaluate sin at each endpoint before applying the u antiderivative. For an indefinite integral, return to x and include C.