Question

How do you solve $\int{\tan xdx}$?

Answer 1

In this problem we have to calculate the integration value of the trigonometric ratio $\tan x$. First, we will convert the given trigonometric ratio into $\sin x$, $\cos x$ by using the basic definitions of trigonometric ratios i.e., $\tan x=\dfrac{\sin x}{\cos x}$. Now we will assume the substitution $u=\cos x$ For this assumption we will calculate the value of $\dfrac{du}{dx}$ by differentiating the equation $u=\cos x$ with respect to $x$. Now we will substitute the values of $u$, $du$ in the integration value and simplify the equation by using the integration formulas. Now we will re substitute the value of $u=\cos x$ after applying the integration formulas and simplify the equation to get the required result.

Complete step by step answer:
Given that, $\int{\tan xdx}$.
From the basic definitions of trigonometric ratios, we have the value of $\tan x$ as $\tan x=\dfrac{\sin x}{\cos x}$. Substituting this value in the above integration value, then we will get
$\Rightarrow \int{\tan xdx}=\int{\dfrac{\sin x}{\cos x}dx}$
By observing the equation, we are going to take the substitution $u=\cos x$ in the above equation. Now differentiating the value of $u$ with respect to $x$, then we will get
$\Rightarrow \dfrac{du}{dx}=\dfrac{d}{dx}\left( \cos x \right)$
We have the differentiation value of $\cos x$ as $-\sin x$, then the above equation modified as
$\Rightarrow \dfrac{du}{dx}=-\sin x$
From the above equation, the value of $du$ will be
$\Rightarrow du=-\sin xdx$
Now substituting the value $u=\cos x$ in the integration value, then we will get
$\Rightarrow \int{\tan x}dx=\int{\dfrac{\sin x}{u}dx}$
Multiplying the above equation with negative signs, then we will have
$\Rightarrow \int{\tan xdx}=-\int{\dfrac{-\sin x}{u}dx}$
Substituting the value $du=-\sin xdx$ in the above equation, then we will get
$\Rightarrow \int{\tan xdx}=-\int{\dfrac{du}{u}}$
We have the integration formula $\int{\dfrac{dx}{x}=\ln x+C}$. Applying this formula in the above equation, then we will get
$\Rightarrow \int{\tan xdx}=-\ln \left| u \right|+C$
Resubstituting the value of $u$ in the above equation, then we will get
$\Rightarrow \int{\tan xdx}=-\ln \left| \cos x \right|+C$
We have the logarithmic formula $-\ln \left| a \right|=\ln \dfrac{1}{a}$. Applying this formula in the above equation, then we will have
$\Rightarrow \int{\tan xdx}=\ln \left( \dfrac{1}{\cos x} \right)+C$
We have the trigonometric formula $\dfrac{1}{\cos x}=\sec x$, then the integration value of the $\tan x$ will be
$\Rightarrow \int{\tan xdx}=\ln \left| \sec x \right|+C$

Note: We can also follow the above procedure to calculate the integral value of $\cot x$. We will use the definition of the $\cot x$ as $\cot x=\dfrac{\cos x}{\sin x}$. We will take the substitution $u=\sin x$ and follow the above procedure to get the integral value of $\cot x$.