Rewriting a hyperbolic function and evaluating its integral

Expert intro: This problem connects hyperbolic identities, algebraic solving, and an integral that becomes simple after rewriting the denominator.

Detailed walkthrough

Part (a): rewrite using exponential definitions

Given

$f(x)=5\cosh x-4\sinh x,$

use the identities

$\cosh x=\frac{e^x+e^{-x}}{2},\qquad \sinh x=\frac{e^x-e^{-x}}{2}.$

Substitute these into $f(x)$:

$f(x)=5\left(\frac{e^x+e^{-x}}{2}\right)-4\left(\frac{e^x-e^{-x}}{2}\right).$

Collect terms:

$f(x)=\frac{1}{2}\bigl(5e^x+5e^{-x}-4e^x+4e^{-x}\bigr)=\frac{1}{2}(e^x+9e^{-x}).$

That proves the required identity.

Part (b): solve $f(x)=5$

Using the exponential form,

$\frac{1}{2}(e^x+9e^{-x})=5.$

Multiply by 2 and then by $e^x$:

$e^x+9e^{-x}=10,$

$e^{2x}+9=10e^x.$

Let $u=e^x$, where $u>0$. Then

$u^2-10u+9=0,$

which factors as

$(u-1)(u-9)=0.$

Hence

$u=1 \quad \text{or} \quad \nu=9.$

So

$x=0 \quad \text{or} \quad x=\ln 9.$

Part (c): evaluate the integral

We want

$\int_{\frac12\ln 3}^{\ln 3}\frac{1}{5\cosh x-4\sinh x}\,dx.$

From part (a), the denominator is

$5\cosh x-4\sinh x=\frac12(e^x+9e^{-x}).$

So the integrand becomes

$\frac{1}{\frac12(e^x+9e^{-x})}=\frac{2}{e^x+9e^{-x}}.$

Multiply top and bottom by $e^x$:

$\frac{2e^x}{e^{2x}+9}.$

Now use the substitution

$u=e^x, \qquad d\nu=e^x\,dx.$

The limits change to

$x=\frac12\ln 3 \Rightarrow u=\sqrt3, \qquad x=\ln 3 \Rightarrow u=3.$

Thus

$\int_{\sqrt3}^{3}\frac{2}{u^2+9}\,du.$

This is a standard arctangent integral:

$\int \frac{2}{u^2+9}\,du=\frac{2}{3}\arctan\left(\frac{u}{3}\right).$

Evaluate:

=\frac{2}{3}\left(\arctan(1)-\arctan\left(\frac{\sqrt3}{3}\right)\right).$$ Since $\arctan(1)=\frac{\pi}{4}$ and $\arctan(\sqrt3/3)=\frac{\pi}{6}$, we get $$\frac{2}{3}\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\frac{2}{3}\cdot\frac{\pi}{12}=\frac{\pi}{18}.$$ ## Key idea to remember Hyperbolic functions often become much easier after converting them to exponentials. In this question, that step turns the equation into a quadratic in $e^x$ and turns the integral into a basic arctangent form. ### 💡 Pitfall guide A typical mistake is to try to integrate $1/(5\cosh x-4\sinh x)$ directly without rewriting it. That leads to unnecessary algebra and often to a wrong antiderivative. Another common error is forgetting that when you set $u=e^x$, the differential is $du=e^x dx$, not just $dx$. Finally, in part (b), some students stop at $e^{2x}-10e^x+9=0$ but forget that $e^x$ must be positive, which is why the substitution $u=e^x$ is so useful and safe. ### 🔄 Real-world variant If the integrand were $\frac{1}{3\cosh x+4\sinh x}$ instead, the same exponential rewrite would produce $\frac{2}{7e^x- e^{-x}}$, leading to a different substitution outcome and potentially a logarithmic form instead of an arctangent. If the equation in part (b) were $5\cosh x-4\sinh x=9$, the same method would give a quadratic in $e^x$ with different roots. The structure of the solution stays the same: rewrite, substitute $u=e^x$, and solve the resulting algebraic equation. ### 🔍 Related terms hyperbolic cosine, exponential substitution, arctangent integral