:::section[Antiderivatives from the Derivative Table] Hyperbolic integration begins by reversing the derivative formulas. The constant of integration must be included, and reciprocal-function formulas require the same domain restrictions as the original functions. ::: :::theorem[Basic Hyperbolic Antiderivatives] On intervals where the integrands are defined, $$\begin{align*} \int \sinh x\,dx&=\cosh x+C,\\ \int \cosh x\,dx&=\sinh x+C,\\ \int \operatorname{sech}^2x\,dx&=\tanh x+C,\\ \int \operatorname{csch}^2x\,dx&=-\coth x+C,\\ \int \operatorname{sech}x\tanh x\,dx&=-\operatorname{sech}x+C,\\ \int \operatorname{csch}x\coth x\,dx&=-\operatorname{csch}x+C. \end{align*}$$ ::: :::proof[Proof] Each identity follows by differentiating the proposed antiderivative. For example, $$\begin{equation*} \frac{d}{dx}\tanh x=\operatorname{sech}^2x \end{equation*}$$ and $$\begin{equation*} \frac{d}{dx}(-\coth x)=\operatorname{csch}^2x. \end{equation*}$$ The remaining formulas follow in the same way. $\square$ ::: :::solved-problem[Linear Substitution] To determine $$\begin{equation*} \int \sinh(4x-3)\,dx, \end{equation*}$$ let $u=4x-3$. Then $du=4\,dx$, so $$\begin{align*} \int \sinh(4x-3)\,dx &=\frac14\int \sinh u\,du\\ &=\frac14\cosh u+C\\ &=\frac14\cosh(4x-3)+C. \end{align*}$$ ::: :::solved-problem[A Quotient Producing a Logarithm] Since $$\begin{equation*} \frac{d}{dx}\cosh x=\sinh x, \end{equation*}$$ one has $$\begin{align*} \int \tanh x\,dx &=\int \frac{\sinh x}{\cosh x}\,dx\\ &=\ln(\cosh x)+C. \end{align*}$$ Because $\cosh x>0$ for every real $x$, no absolute value is needed. ::: :::key-formula[Useful Logarithmic Antiderivatives] On intervals where the integrands are defined, $$\begin{align*} \int \coth x\,dx&=\ln|\sinh x|+C,\\ \int \tanh x\,dx&=\ln(\cosh x)+C,\\ \int \operatorname{sech}x\,dx&=\tan^{-1}(\sinh x)+C,\\ \int \operatorname{csch}x\,dx&=\ln\left|\tanh\frac{x}{2}\right|+C. \end{align*}$$ ::: :::proof[Verification of the Secant Formula] Using the chain rule, $$\begin{align*} \frac{d}{dx}\tan^{-1}(\sinh x) &=\frac{\cosh x}{1+\sinh^2x}\\ &=\frac{\cosh x}{\cosh^2x}\\ &=\operatorname{sech}x. \end{align*}$$ Thus the stated formula is valid. $\square$ ::: :::mistake[Do Not Forget the Interval] The formula for $\int\coth x\,dx$ is used separately on $(-\infty,0)$ and $(0,\infty)$ because $\coth x$ is undefined at $x=0$. ::: :::exercise[Practice Questions] 1. Evaluate $\int \cosh(5x)\,dx$. 2. Evaluate $\int \operatorname{sech}^2(3x-1)\,dx$. 3. Evaluate $\int \coth(2x)\,dx$ on an interval not containing zero. 4. Verify by differentiation that $\int \operatorname{sech}x\,dx=\tan^{-1}(\sinh x)+C$. ::: :::answer[Answers and Guidance] 1. $\frac15\sinh(5x)+C$. 2. $\frac13\tanh(3x-1)+C$. 3. $\frac12\ln|\sinh(2x)|+C$. 4. Differentiate and use $1+\sinh^2x=\cosh^2x$. ::: :::faq[Frequently Asked Questions] Q: Why is there no absolute value in $\ln(\cosh x)$? A: Because $\cosh x\ge1$ for every real $x$. Q: Why does $\int \operatorname{csch}^2x\,dx$ have a minus sign? A: Because $(\coth x)'=-\operatorname{csch}^2x$. Q: Can one integrate a composite hyperbolic function without substitution? A: Only when the derivative of the inner function is already present up to a constant factor. ::: :::call-to-action[Continue Learning] subtitle: Recognize radical and rational integrals whose antiderivatives are inverse hyperbolic functions. button: Study Standard Inverse-Hyperbolic Integrals|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions/standard-integrals-with-inverse-hyperbolic-forms button-ghost: Review This Section|/real-analysis/functions/hyperbolic-and-inverse-hyperbolic-functions :::