Problem 160
Evaluate the integral: I = integral x sinh x dx
Evaluate the integral: $$ I = \int x\sinh x\,dx. $$
Explanation of Method
Integration by parts with $u=x$, $dv=\sinh x\,dx$ leaves a leftover integral of $\cosh x$ alone, which is elementary.
- Choose $u=x$, $dv=\sinh x\,dx$.
- Differentiate $u$ and integrate $dv$.
- Apply the integration by parts formula and integrate the remainder.
Step 1 · Set up integration by parts
Let:
$$ u = x \quad \Rightarrow \quad du = dx, \qquad dv = \sinh x\,dx \quad \Rightarrow \quad v = \cosh x. $$
Step 2 · Apply the formula
$$ I = x\cosh x - \int \cosh x\,dx. $$
Step 3 · Integrate
Since $\int \cosh x\,dx = \sinh x$:
$$ I = x\cosh x - \sinh x + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$ | $x$ |
| $dv$ | $\sinh x\,dx$ |
| Antiderivative | $x\cosh x-\sinh x$ |
Additional Notes
- The integrand is defined for all real $x$.
- Unlike the circular case ($\int x\sin x\,dx$ from Problem 158), no alternating sign appears here because both $\int\sinh x\,dx=\cosh x$ and $\int\cosh x\,dx=\sinh x$ carry a plus sign — the hyperbolic functions don't introduce the sign flips that circular trig functions do.
- The result can be checked by differentiating $x\cosh x-\sinh x$ and confirming it returns $x\sinh x$.
Result
The solution to the integral is:
$$
I = x\cosh x - \sinh x + C.
$$