FP3.4 - Integration

Hyperbolic antiderivatives reverse the derivative patterns from FP3.3. First recognise whether the integrand contains a function together with the derivative of its argument; then account for any constant chain factor. Always add $C$ to an indefinite integral.

Ratios lead to logarithms because $d(\ln|f|)/dx=f'/f$:$$\int\tanh x\,dx=\ln(\cosh x)+C,$$$$\int\coth x\,dx=\ln|\sinh x|+C.$$The absolute value is needed in the second result because $\sinh x$ may be negative; $\cosh x$ is always positive.

Two useful single-function forms follow by differentiating the inverse expression:$$\int\operatorname{sech}x\,dx=\arctan(\sinh x)+C,$$$$\int\operatorname{cosech}x\,dx=\ln\left|\tanh\frac{x}{2}\right|+C.$$These apply only on intervals where the integrand is defined.

For$$\int\left(3\cosh(3x)-2\operatorname{sech}^2(2x)\right)dx,$$each inner derivative is already present, so the result is$$\sinh(3x)-\tanh(2x)+C.$$Differentiate the result: $3\cosh(3x)-2\operatorname{sech}^2(2x)$ is recovered exactly.

Expressions involving powers may need an identity before integration. For example,$$\cosh^2x=\frac{1+\cosh2x}{2},\qquad \sinh^2x=\frac{\cosh2x-1}{2}.$$Thus $\int\cosh^2x\,dx=x/2+\sinh2x/4+C$. Choose an identity that turns the integrand into direct patterns rather than expanding exponentials unnecessarily.

An inverse trigonometric or inverse hyperbolic function is usually integrated by parts. Choose the inverse function as the part to differentiate and $1$ as the part to integrate:$$u=f^{-1}(x),\qquad dv=dx.$$Its derivative becomes an algebraic expression that can be simplified or integrated directly.

For example,$$\int\operatorname{arsinh}x\,dx=x\operatorname{arsinh}x-\int\frac{x}{\sqrt{1+x^2}}\,dx.$$With $w=1+x^2$, the remaining integral is $\sqrt{1+x^2}$, giving$$x\operatorname{arsinh}x-\sqrt{1+x^2}+C.$$Differentiating this result makes the two $x/\sqrt{1+x^2}$ terms cancel.

For a scaled argument, substitute first or apply integration by parts with the full argument. If $a>0$, then$$\int\arctan\left(\frac{x}{a}\right)dx=x\arctan\left(\frac{x}{a}\right)-\frac a2\ln(a^2+x^2)+C.$$Terms differing only by a constant inside a logarithm are absorbed into $C$.

For a definite integral, use one antiderivative consistently and retain exact principal inverse values. Check that the complete interval lies in the real domain: $[-1,1]$ for $\arcsin x$, $[1,\infty)$ for $\operatorname{arcosh}x$, and $(-1,1)$ for $\operatorname{artanh}x$.

A substitution is chosen so that a standard identity removes a quadratic expression. Assume $a>0$. Trigonometric substitutions are natural for $a^2-x^2$ and $a^2+x^2$; hyperbolic substitutions give equally direct forms for positive sums and differences. Transform the differential and limits as well as the integrand.

Use the same four moves each time: choose the identity-matching substitution; calculate $dx$; replace every occurrence of $x$; simplify the square root with the parameter range that fixes its sign. For a definite integral, convert both bounds immediately and evaluate in the new variable without back-substituting.

For $0<b<a$,$$\int_0^b\frac{dx}{(a^2-x^2)^{3/2}}$$uses $x=a\sin\theta$, $dx=a\cos\theta\,d\theta$, with $0\le\theta<\pi/2$. Since $(a^2-x^2)^{3/2}=a^3\cos^3\theta$, the integral becomes$$\frac1{a^2}\int_0^{\arcsin(b/a)}\sec^2\theta\,d\theta=\frac{b}{a^2\sqrt{a^2-b^2}}.$$The range justifies replacing $\sqrt{\cos^2\theta}$ by $\cos\theta$.

A hyperbolic substitution can be especially clean when $\sqrt{a^2+x^2}$ or $\sqrt{x^2-a^2}$ appears because the radical becomes $a\cosh t$ or $a\sinh t$. A trigonometric route may instead be preferable when exact angle values make definite limits simple. The valid method is the one whose identity matches the sign pattern and whose range is controlled.

For an integral involving $\sqrt{Ax^2+Bx+C}$, expose the quadratic's geometry before substituting. Complete the square, factor out any positive constant, and identify one of the patterns $a^2-u^2$, $a^2+u^2$ or $u^2-a^2$. In more complicated cases the required substitution may be supplied, but every algebraic change must still be shown.

Consider$$\int\frac{dx}{\sqrt{x^2+6x+13}}.$$Completing the square gives $(x+3)^2+4$. Put $x+3=2\sinh t$, so $dx=2\cosh t\,dt$ and the denominator is $2\cosh t$. The integral is $\int dt$, hence$$\operatorname{arsinh}\left(\frac{x+3}{2}\right)+C.$$Differentiation recovers the original integrand.

If the numerator contains the derivative $2Ax+B$ of the quadratic, try the direct substitution $u=Ax^2+Bx+C$ before a trigonometric or hyperbolic substitution. For example,$$\int\frac{2x+4}{\sqrt{x^2+4x+7}}\,dx=2\sqrt{x^2+4x+7}+C.$$This is shorter because the numerator already supplies $du$.

When a non-obvious substitution is given, derive its consequences rather than guessing the target. Calculate $dx$, rewrite the surd, state the parameter range needed for its sign, and simplify systematically. For definite integrals, translate endpoints before cancellation so that orientation and sign remain visible.

A reduction formula rewrites an integral indexed by $n$ in terms of the same family with a smaller index. It is derived from the definition of the indexed integral, usually by integration by parts and an identity. State the definition, restrictions and base cases before using the recurrence.

Let$$I_n=\int_0^{\pi/2}\sin^n x\,dx.$$For $n\ge2$, write $\sin^n x=\sin^{n-1}x\sin x$ and integrate by parts with $u=\sin^{n-1}x$, $dv=\sin x\,dx$. The boundary term $[-\sin^{n-1}x\cos x]_0^{\pi/2}$ is zero.

The remaining integral is$$(n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2x\,dx=(n-1)(I_{n-2}-I_n).$$Therefore$$nI_n=(n-1)I_{n-2},\qquad n\ge2.$$Use $I_0=\pi/2$ for even powers and $I_1=1$ for odd powers. For instance, $I_4=(3/4)I_2=(3/4)(1/2)I_0=3\pi/16$.

A different family needs a different derivation. If$$J_n=\int\frac{\sin nx}{\sin x}\,dx,\qquad n>0,$$then $\sin((n+2)x)-\sin(nx)=2\cos((n+1)x)\sin x$. Dividing by $\sin x$ and integrating gives$$J_{n+2}=\frac{2\sin((n+1)x)}{n+1}+J_n.$$This is an indefinite relation, so use a consistent arbitrary constant convention.

For an unfamiliar indexed family, choose integration by parts so that one factor differentiates toward the lower-index integrand. Rewrite leftover powers using an identity, isolate every occurrence of the original $I_n$, and only then divide to obtain the recurrence. To evaluate, step down repeatedly until a directly calculable base integral is reached.

Arc length adds infinitesimal distances $ds$ along a curve. A surface of revolution adds circumference times arc length, so the radius is the perpendicular distance from the curve to the axis. The specification covers Cartesian and parametric curves, not polar forms.

First differentiate and simplify the speed factor under the square root. Then determine the radius and parameter interval. Use absolute values or split the interval if the signed coordinate crosses the axis. The square root denotes non-negative speed, so $\sqrt{h(t)^2}=|h(t)|$ unless its sign is established.

For$$x=t-\frac{t^3}{3},\qquad y=t^2,\qquad 0\le t\le1,$$we have $dx/dt=1-t^2$ and $dy/dt=2t$, so$$\frac{ds}{dt}=\sqrt{(1-t^2)^2+4t^2}=1+t^2.$$Hence the arc length is $[t+t^3/3]_0^1=4/3$. Rotating the curve about the $x$-axis gives curved area$2\pi\int_0^1t^2(1+t^2)dt=\frac{16\pi}{15}.

The formula $2\pi\int r\,ds$ gives only the curved surface swept out by the curve. If a solid's total surface area is requested and the endpoints generate circular end faces, add the appropriate disc or annulus areas separately. Read the geometry and endpoint radii before deciding whether end faces exist.

Length has units of length and surface area has squared units. Both must be non-negative. For a definite calculation, verify that the parameter traces the intended piece once; retracing a segment would count its length or swept area again.