Evaluate the integral: integral (m sinh x + n cosh x) dx

Evaluate the integral: $$ \int (m \sinh x + n \cosh x) \, dx, $$ where $m$ and $n$ are constants.

Step 1 · Use the standard integrals of $\sinh x$ and $\cosh x$

Recall the following standard integrals:

$$ \int \sinh x\,dx = \cosh x \qquad \text{and} \qquad \int \cosh x\,dx = \sinh x. $$

Using these, we can break the integral into two parts.

Step 2 · Split the integral

We can now split the given integral as follows:

$$ \int (m\sinh x + n\cosh x)\,dx = m\int \sinh x\,dx + n\int \cosh x\,dx. $$

Step 3 · Solve each integral

Now, we integrate each term:

$$ m\int \sinh x\,dx = m\cosh x $$

and

$$ n\int \cosh x\,dx = n\sinh x. $$

Step 4 · Combine the results

Combining the results from the two integrals, we get:

$$ \int (m\sinh x + n\cosh x)\,dx = m\cosh x + n\sinh x + C, $$

where $C$ is the constant of integration.

Summary Table

Additional Notes

• The integrand is defined for all real $x$, since $\sinh x$ and $\cosh x$ are defined everywhere.
• Unlike ordinary trig functions where $\int \sin x\,dx = -\cos x$, hyperbolic functions swap into each other without a sign change: $\int \sinh x\,dx=\cosh x$ and $\int \cosh x\,dx=\sinh x$ — a useful distinction to remember.
• The result can be checked by differentiating $m\cosh x+n\sinh x$ and confirming it returns $m\sinh x+n\cosh x$, using $\frac{d}{dx}\cosh x=\sinh x$ and $\frac{d}{dx}\sinh x=\cosh x$.