Question

How do you find the integral of ${\tan ^2}\left( x \right)\sec \left( x \right)dx$?

Answer 1

Hint:
First, use trigonometry identity to find the value of ${\tan ^2}x$ and put the value of ${\tan ^2}x$ in the given integral. Then, use distributive property in integral. Next, use property (III) to split the integrals. The, solve the first integral by integration by parts and second integral using (IV). Substitute all the values in the combined integral and get the required result.

Formula used:
1) The integral of the product of a constant and a function = the constant $ \times $ integral of the function.
i.e., $\int {\left( {kf\left( x \right)dx} \right)} = k\int {f\left( x \right)dx} $, where $k$ is a constant.
2) Trigonometric identity: ${\tan ^2}x + 1 = {\sec ^2}x$
3) The integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} $
4) Integration formula: $\int {{{\sec }^2}xdx} = \tan x$ and \[\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right|\]
5) Differentiation formula: $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
6) Integration by parts: $\int {udv} = uv - \int {vdu} $

Complete step by step solution:
We have to find $\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} $…(i)
Use identity ${\tan ^2}x - {\sec ^2}x = 1$, to find the value of ${\tan ^2}x$.
Since, ${\tan ^2}x + 1 = {\sec ^2}x$.
So, ${\tan ^2}x = {\sec ^2}x - 1$.
Now, put the value of ${\tan ^2}x$ in integral (i).
$\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \int {\left( {{{\sec }^2}x - 1} \right)\sec xdx} $…(ii)
Use distributive property in integral (ii).
$\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \int {\left( {{{\sec }^3}x - \sec x} \right)dx} $…(iii)
Now, using the property that the integral of the sum or difference of a finite number of functions is equal to the sum or difference of the integrals of the various functions.
i.e., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx} $
So, in above integral (iii), we can use above property
$\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \int {{{\sec }^3}xdx} - \int {\sec xdx} $…(iv)
First integral: $\int {{{\sec }^3}xdx} $
It can be written as $\int {{{\sec }^2}x\sec xdx} $
Now, use integration by parts with $u = \sec x$ and $dv = {\sec ^2}xdx$…(v)
Differentiate $u$ with respect to $x$.
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {\sec x} \right)$…(vi)
Now, using the differentiation formula $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$ in differentiation (vi), we get
$\dfrac{{du}}{{dx}} = \sec x\tan x$
$ \Rightarrow du = \sec x\tan xdx$…(vii)
Now, integrate $v$ with respect to $x$.
$\int {dv} = \int {{{\sec }^2}xdx} $…(viii)
Now, using the integration formula $\int {{{\sec }^2}xdx} = \tan x$ in integral (viii), we get
$v = \tan x$…(ix)
The integration by parts formula is:
$\int {udv} = uv - \int {vdu} $
Put the value of $u,v,du,dv$ from (v), (vii) and (ix).
$\int {{{\sec }^2}x\sec xdx} = \sec x\tan x - \int {\tan x\sec x\tan xdx} $
$ \Rightarrow \int {{{\sec }^2}x\sec xdx} = \sec x\tan x - \int {{{\tan }^2}x\sec xdx} $…(x)
Second integral: $\int {\sec xdx} $
Now, using the integration formula \[\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right|\] in second integral, we get
\[ \Rightarrow \int {\sec xdx} = \ln \left| {\sec x + \tan x} \right|\]…(xi)
Use equation (x) and (xi) in equation (iv).
$\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \sec x\tan x - \int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} + \ln \left| {\sec x + \tan x} \right|$
We can add the $\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} $ on the right to the left to get:
$2\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \sec x\tan x + \ln \left| {\sec x + \tan x} \right|$
$ \Rightarrow \int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \dfrac{1}{2}\sec x\tan x + \dfrac{1}{2}\ln \left| {\sec x + \tan x} \right| + C$

Hence, $\int {{{\tan }^2}\left( x \right)\sec \left( x \right)dx} = \dfrac{1}{2}\sec x\tan x + \dfrac{1}{2}\ln \left| {\sec x + \tan x} \right| + C$.

Note:
If instead we had $\int {\tan x{{\sec }^2}xdx} $, we could make the simple substitution $u = \tan x$, $du = {\sec ^2}xdx$, which would make the integral $\int {udu} $. Unfortunately, that is not the case.