Question

Evaluate the integral: $\int{\cos x\log \left( \tan \dfrac{x}{2} \right)dx}$

Answer 1

To integrate the given integration using integration by parts, we are going to use the following integration by parts formula which is equal to: $\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx}$. The catch in this integration is what could be the first term and what could be the second term. And this priority of the function is decided by the mnemonics named “ILATE” and we will discuss each letter in the solution below.

Complete step-by-step solution:
The integration which we have to find is as follows:
$\int{\cos x\log \left( \tan \dfrac{x}{2} \right)dx}$
Now, we are going to integrate by using integration by parts. The formula for integration by parts is as follows:
$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx}-\int{f'\left( x \right)}\int{g\left( x \right)dx}$ …………… (1)
In the above formula, $f\left( x \right)\And g\left( x \right)$ are two functions but here, we should know which function will get the priority over another. To decide the priority there is mnemonic which is equal to:
ILATE
Now, in the above mnemonic, “I” represents inverse functions, “L” represents a logarithmic function, “A” represents algebraic functions, “T” represents trigonometric functions and “E” represents exponential functions. Now, the order of the letters in this word “ILATE” will be the order of priority for the functions.
So, using this mnemonic in the given integration, in the given integration, we have two functions, logarithmic and trigonometric functions, so logarithm will get priority over trigonometric functions. Now, $f\left( x \right)=\log \left( \tan \dfrac{x}{2} \right)\And g\left( x \right)=\cos x$ then substituting these values of f(x) and g(x) in eq. (1) we get,
$\int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\int{\cos dx}-\int{\left( \log \left( \tan \dfrac{x}{2} \right) \right)'}\int{\cos xdx}$ ……………. (2)
We know the integration of $\cos x$ with respect to x we get,
$\int{\cos xdx}=\sin x$
$\int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-\int{\left( \log \left( \tan \dfrac{x}{2} \right) \right)'}\left( \sin x \right)dx$…………(3)
Now, we are going to show the derivative of $\log \left( \tan \dfrac{x}{2} \right)$ with respect to x as follows:
$\dfrac{d\log \left( \tan \dfrac{x}{2} \right)}{dx}=\dfrac{1}{\tan \dfrac{x}{2}}\left( {{\sec }^{2}}\dfrac{x}{2} \right)\left( \dfrac{1}{2} \right)$
We know that $\tan \dfrac{x}{2}=\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}\And \sec \dfrac{x}{2}=\dfrac{1}{\cos \dfrac{x}{2}}$ so using the relation in the above we get,
$\begin{aligned} & \dfrac{d\log \left( \tan \dfrac{x}{2} \right)}{dx}=\dfrac{1}{\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}\left( \dfrac{1}{{{\cos }^{2}}\dfrac{x}{2}} \right)\left( \dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{d\log \left( \tan \dfrac{x}{2} \right)}{dx}=\dfrac{1}{\dfrac{\sin \dfrac{x}{2}}{1}}\left( \dfrac{1}{\cos \dfrac{x}{2}} \right)\left( \dfrac{1}{2} \right) \\\ & \Rightarrow \dfrac{d\log \left( \tan \dfrac{x}{2} \right)}{dx}=\dfrac{1}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \\\ \end{aligned}$
We know the identity of sine of double angle as follows:
$\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$
Using the above relation in the above equation we get,
$\Rightarrow \dfrac{d\log \left( \tan \dfrac{x}{2} \right)}{dx}=\dfrac{1}{\sin x}$
Using the above relation in (3) we get,
$\begin{aligned} & \int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-\int{\left( \log \left( \tan \dfrac{x}{2} \right) \right)'}\left( \sin x \right)dx \\\ & \Rightarrow \int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-\int{\left( \dfrac{1}{\sin x} \right)}\left( \sin x \right)dx \\\ & \Rightarrow \int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-\int{dx} \\\ & \Rightarrow \int{\log \left( \tan \dfrac{x}{2} \right)\cos xdx}=\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-x+C \\\ \end{aligned}$
From the above solution, we have shown the integration of $\int{\cos x\log \left( \tan \dfrac{x}{2} \right)dx}$ which is equal to $\log \left( \tan \dfrac{x}{2} \right)\left( \sin x \right)-x+C$.

Note: The point of making a mistake is in the integration of $\sin x\And \cos x$ and the mistake is in putting the sign before the result of integration.
In the integration of $\cos x$ with respect to x we get a negative sign in the result whereas in the integration of $\sin x$ with respect to x we get a positive sign in the result.
$\begin{aligned} & \int{\cos dx}=\sin x \\\ & \int{\sin xdx}=-\cos x \\\ \end{aligned}$