Problem 161
Evaluate the integral: I = integral x cosh x dx
Evaluate the integral: $$ I = \int x\cosh x\,dx. $$
Explanation of Method
Integration by parts with $u=x$, $dv=\cosh x\,dx$ leaves a leftover integral of $\sinh x$ alone, which is elementary.
- Choose $u=x$, $dv=\cosh x\,dx$.
- Differentiate $u$ and integrate $dv$.
- Apply the integration by parts formula and integrate the remainder.
Step 1 · Set up integration by parts
Let:
$$ u = x \quad \Rightarrow \quad du = dx, \qquad dv = \cosh x\,dx \quad \Rightarrow \quad v = \sinh x. $$
Step 2 · Apply the formula
$$ I = x\sinh x - \int \sinh x\,dx. $$
Step 3 · Integrate
Since $\int \sinh x\,dx = \cosh x$:
$$ I = x\sinh x - \cosh x + C. $$
Summary Table
| Quantity | Value |
|---|---|
| $u$ | $x$ |
| $dv$ | $\cosh x\,dx$ |
| Antiderivative | $x\sinh x-\cosh x$ |
Additional Notes
- The integrand is defined for all real $x$.
- This is the counterpart of Problem 160 with $\sinh x$ and $\cosh x$ swapped; like the hyperbolic sine case, no alternating sign appears since both hyperbolic integrals ($\int\sinh x\,dx=\cosh x$, $\int\cosh x\,dx=\sinh x$) carry a plus sign.
- The result can be checked by differentiating $x\sinh x-\cosh x$ and confirming it returns $x\cosh x$.
Result
The solution to the integral is:
$$
I = x\sinh x - \cosh x + C.
$$