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# How do you find the integral of $\left( \cos x \right)\left( \cosh x \right)dx$?

Last updated date: 23rd Jun 2024
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Hint: In this problem we have given a function and asked to calculate the integration value. We can observe that the given function is the multiplication of the trigonometric function with the hyperbolic function. First of all, the given function has multiplication operation, so we are going to use the integration by parts rule which is $\int{uvdx}=u\int{vdx}-\int{\left[ \left( {{u}^{'}} \right)\int{vdx} \right]dx}$. So, we will choose $u$, $v$ by ILATE rule. After choosing $u$, $v$. We will apply the integration by parts rule and simplify the equation by using the integration and differentiation formulas.

Given that, $\left( \cos x \right)\left( \cosh x \right)dx$.
In the above function we have trigonometric function $\cos x$, hyperbolic function $\cosh x$. From ILATE (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential) rule the first function $u$ is $\cos x$, the second function $v$ is $\cosh x$.
Now the integration of the given function from the integration by parts formula $\int{uvdx}=u\int{vdx}-\int{\left[ \left( {{u}^{'}} \right)\int{vdx} \right]dx}$ is given by
$\int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ {{\left( \cos x \right)}^{'}}\int{\cosh xdx} \right]dx}$
We have the differentiation formula $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, applying this formula in the above equation, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\int{\cosh xdx}-\int{\left[ \left( -\sin x \right)\int{\cosh xdx} \right]dx}$
We have the integration formula $\int{\cosh xdx}=\sinh x+C$, applying this formula in the above equation, then we will have
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\left( \sinh x \right)+\int{\sin x\sinh xdx}...\left( \text{i} \right)$
From the above equation we can say that to calculate the integration of the given function we need to have the value of $\int{\sin x\sinh xdx}$. So, applying the same integration by parts formula for this function also, then we will have
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\int{\sinh xdx}-\int{\left[ {{\left( \sin x \right)}^{'}}\int{\sinh xdx} \right]dx}$
We have the formulas $\dfrac{d}{dx}\left( \sin x \right)=\cos x$, $\int{\sinh xdx}=\cosh x+C$. Substituting these formulas in the above equation, then we will get
$\Rightarrow \int{\sin x\sinh xdx}=\sin x\cosh x-\int{\cos x\cosh xdx}...\left( \text{ii} \right)$
Substituting this value in the equation $\left( \text{i} \right)$, then we will get
$\Rightarrow \int{\cos x\cosh xdx}=\cos x\sinh x+\left[ \sin x\cosh x-\int{\cos x\cosh xdx} \right]$
Simplifying the above equation, then we will have
\begin{align} & \Rightarrow \int{\cos x\cosh xdx}+\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\ & \Rightarrow 2\int{\cos x\cosh xdx}=\cos x\sinh x+\sin x\cosh x \\ & \therefore \int{\cos x\cosh xdx}=\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C \\ \end{align}

Hence the integration of the given function $\left( \cos x \right)\left( \cosh x \right)dx$ is $\dfrac{\cos x\sinh x+\sin x\cosh x}{2}+C$.

Note: In this problem students may make a little mistake which will cost the whole problem. In integration of trigonometric ratios, we have $\int{\sin xdx}=-\cos x+C$, $\int{\cos xdx}=\sin x+C$ but when it comes to integration of hyperbolic functions, we have $\int{\sinh xdx}=\cosh x+C$, $\int{\cosh xdx}=\sinh x+C$. If you forget about this change and solved the problem you will get an incorrect solution.