:::section[Choosing an Integration Strategy] Hyperbolic integrals are simplified by substitution when one derivative factor is present, by identities when powers occur, and by convergence analysis when an endpoint is infinite. ::: :::solved-problem[Power of Hyperbolic Sine] To determine $$\begin{equation*} \int \sinh^6x\cosh x\,dx, \end{equation*}$$ let $u=\sinh x$. Then $du=\cosh x\,dx$, and $$\begin{equation*} \int \sinh^6x\cosh x\,dx =\int u^6\,du =\frac{u^7}{7}+C =\frac{\sinh^7x}{7}+C. \end{equation*}$$ ::: :::solved-problem[Cotangent and Cosecant Powers] Since $$\begin{equation*} \frac{d}{dx}\coth x=-\operatorname{csch}^2x, \end{equation*}$$ let $u=\coth x$: $$\begin{align*} \int \coth^2x\,\operatorname{csch}^2x\,dx &=-\int u^2\,du\\ &=-\frac13u^3+C\\ &=-\frac13\coth^3x+C. \end{align*}$$ ::: :::solved-problem[Definite Integral] For $a\ne0$, $$\begin{align*} \int_0^{45}\cosh\left(\frac{x}{a}\right)\,dx &=a\left[\sinh\left(\frac{x}{a}\right)\right]_0^{45}\\ &=a\sinh\left(\frac{45}{a}\right). \end{align*}$$ ::: :::theorem[Improper Integral of Hyperbolic Secant] $$\begin{equation*} \int_0^\infty \operatorname{sech}x\,dx=\frac{\pi}{2}. \end{equation*}$$ ::: :::proof[Proof] Using $\int\operatorname{sech}x\,dx=\tan^{-1}(\sinh x)+C$, $$\begin{align*} \int_0^\infty\operatorname{sech}x\,dx &=\lim_{b\to\infty}\left[\tan^{-1}(\sinh x)\right]_0^b\\ &=\frac{\pi}{2}-0\\ &=\frac{\pi}{2}. \end{align*}$$ The limit exists because $\sinh b\to\infty$. $\square$ ::: :::solved-problem[Half-Angle Denominator] Using $1+\cosh x=2\cosh^2(x/2)$, $$\begin{align*} \int\frac{dx}{1+\cosh x} &=\frac12\int\operatorname{sech}^2\left(\frac{x}{2}\right)\,dx\\ &=\tanh\left(\frac{x}{2}\right)+C. \end{align*}$$ ::: :::theorem[Fourth Power of Hyperbolic Sine] $$\begin{equation*} \int\sinh^4x\,dx =\frac1{32}\sinh4x-\frac14\sinh2x+\frac38x+C. \end{equation*}$$ ::: :::proof[Proof] Using $\sinh^2x=(\cosh2x-1)/2$, $$\begin{align*} \sinh^4x &=\frac14(\cosh2x-1)^2\\ &=\frac18\cosh4x-\frac12\cosh2x+\frac38. \end{align*}$$ Integrating term by term gives the formula. $\square$ ::: :::mistake[Check Improper Endpoints Before Substitution] An antiderivative alone does not prove convergence. For an improper integral, first replace the infinite endpoint by a finite parameter, evaluate, and then take the limit. ::: :::exercise[Practice Questions] 1. Evaluate $\int\cosh^3x\,dx$. 2. Evaluate $\int_0^1x\,\operatorname{sech}^2(x^2)\,dx$. 3. Evaluate $\int\sinh^2x\cosh^2x\,dx$. 4. Determine whether $\int_1^\infty\operatorname{csch}x\,dx$ converges and evaluate it. ::: :::answer[Answers and Guidance] 1. Write $\cosh^3x=\cosh x(1+\sinh^2x)$ and use $u=\sinh x$: $\sinh x+\frac13\sinh^3x+C$. 2. Let $u=x^2$: $\frac12\tanh1$. 3. Use $\sinh^2x\cosh^2x=\frac18(\cosh4x-1)$: $\frac1{32}\sinh4x-\frac18x+C$. 4. It converges and equals $-\ln(\tanh(1/2))$. ::: :::faq[Frequently Asked Questions] Q: When should a power be reduced by identities? A: Use identities when no single derivative factor is available for a direct substitution. Q: Why is $\int_0^\infty\operatorname{sech}x\,dx$ finite? A: Hyperbolic secant decays exponentially as $x\to\infty$. Q: How can a definite hyperbolic integral be checked? A: Differentiate the antiderivative and verify the endpoint substitution separately. ::: :::call-to-action[Continue Learning] subtitle: Review the full sequence from exponential construction through inverse functions, differentiation, and integration. button: Review the Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions button-ghost: Return to Real Functions|/real-analysis/functions :::