2.4 Integration

Use cosh²x−sinh²x=1, 1−tanh²x=sech²x and coth²x−1=csch²x to replace a difficult power or quotient by a form with a known derivative.

Choose the identity that leaves a differential factor. For even powers, separate one factor when integrating; for equations, isolate a single hyperbolic function before applying an inverse.

∫sinh²x dx can use sinh²x=(cosh2x−1)/2, giving a sum of an elementary term and a linear term in x.

The sign in cosh²−sinh² is opposite to the circular identity, and applying an identity without preserving the differential can make the integral harder.

A reduction formula expresses an integral I_n in terms of I_{n−1} or I_{n−2}, usually by integration by parts. Repeating the relation eventually reaches a base integral that can be evaluated directly.

Define the indexed integral and its limits clearly, derive the recurrence once, then apply it with the correct starting value. Track boundary terms and signs at each step.

For I_n=∫sin^n x dx on a fixed interval, integration by parts can relate I_n to I_{n−2}; repeated use reduces an even or odd power to I_0 or I_1.

A recurrence is not the final numerical answer: the base case and the index range are part of the proof.

The signed area between y=f(x) and the x-axis from a to b is ∫_a^b f(x)dx. Geometric area requires splitting where f changes sign or integrating top minus bottom between curves.

Find intersections first and use the interval that traces the intended region once. In parametric or polar forms, change the area formula rather than forcing Cartesian dx.

If f is below the x-axis on [a,b], the geometric area is −∫_a^b f(x)dx. For two curves, integrate (upper−lower) only after checking which is larger throughout.

A definite integral can be zero while geometric area is positive because positive and negative signed regions cancel.

For y=f(x), arc length from a to b is ∫√(1+(dy/dx)²)dx. Rotating the curve about an axis gives surface area 2π∫(radius)×(arc-length element), with the radius and variable chosen consistently.

Parametric and polar curves require their own ds formula. Find endpoints and avoid counting a self-intersecting curve or using signed radius without considering the geometry.

For y=x² from 0 to 1, ds=√(1+4x²)dx; surface area about the x-axis uses radius y=x², so the integrand is 2πx²√(1+4x²).

Arc length is not ∫dy or ∫|f|dx, and surface area is not volume; use the local slant factor.