Question

How to derive power reducing formula for $\int {\left( {{{\sec }^n}x} \right)dx}$ and $\int {\left( {{{\tan }^n}x} \right)dx}$ for integration?

Answer 1

We have to derive power reducing formula for $\int {\left( {{{\sec }^n}x} \right)dx}$ and $\int {\left( {{{\tan }^n}x} \right)dx}$ for integration. For $\int {\left( {{{\sec }^n}x} \right)dx}$, began by writing ${\sec ^n}x$ as ${\sec ^{n - 2}}x \cdot {\sec ^2}xdx$ and letting $u = {\sec ^{n - 2}}x$ and $dv = {\sec ^2}xdx$. Then use integration by parts to derive the result. For $\int {\left( {{{\tan }^n}x} \right)dx}$, began by writing ${\tan ^n}x$ as ${\tan ^{n - 2}}x \cdot {\tan ^2}x$ and then use trigonometry identity to convert ${\tan ^2}x$ into ${\sec ^2}x$. Then, let $u = {\tan ^{n - 2}}x$ and $dv = {\sec ^2}xdx$ in integration by parts and derive the result.
Formula used:
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.
Trigonometric identity: ${\tan ^2}x + 1 = {\sec ^2}x$
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., $\int {\left[ {f\left( x \right) \pm g\left( x \right)} \right]dx} = \int {f\left( x \right)dx} \pm \int {g\left( x \right)dx}$
Integration formula: $\int {{{\sec }^2}xdx} = \tan x$ and $\int {\sec xdx} = \ln \left| {\sec x + \tan x} \right|$
Differentiation formula: $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
Integration by parts: $\int {udv} = uv - \int {vdu}$

Complete step by step solution:
We have to derive power reducing formula for $\int {\left( {{{\sec }^n}x} \right)dx}$ and $\int {\left( {{{\tan }^n}x} \right)dx}$ for integration.
Consider $\int {\left( {{{\sec }^n}x} \right)dx}$
It can be written as $\int {{{\sec }^{n - 2}}x \cdot {{\sec }^2}xdx}$
Now, use integration by parts with $u = {\sec ^{n - 2}}x$ and $dv = {\sec ^2}xdx$…(i)
Differentiate $u$ with respect to $x$.
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\sec }^{n - 2}}x} \right)$…(ii)
Now, using the differentiation formula $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$ in differentiation (ii), we get
$\dfrac{{du}}{{dx}} = \left( {n - 2} \right){\sec ^{n - 2}}x\left( {\sec x\tan x} \right)$
$\Rightarrow du = \left( {n - 2} \right){\sec ^{n - 1}}x\tan x$…(iii)
Now, integrate $v$ with respect to $x$.
$\int {dv} = \int {{{\sec }^2}xdx}$…(iv)
Now, using the integration formula $\int {{{\sec }^2}xdx} = \tan x$ in integral (iv), we get
$v = \tan x$…(v)
The integration by parts formula is:
$\int {udv} = uv - \int {vdu}$
Put the value of $u,v,du,dv$ from (i), (iii) and (v).
$\int {{{\sec }^{n - 2}}x \cdot {{\sec }^2}xdx} = {\sec ^{n - 2}}x\tan x - \int {\tan x\left( {n - 2} \right){{\sec }^{n - 1}}x\tan xdx}$
Using property that 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.
$\Rightarrow \int {{{\sec }^n}xdx} = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {{{\tan }^2}x{{\sec }^{n - 1}}xdx}$…(vi)
Now, put the value of ${\tan ^2}x$ in integral (vi).
$\Rightarrow \int {{{\sec }^n}xdx} = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {\left( {{{\sec }^x} - 1} \right){{\sec }^{n - 1}}xdx}$
$\Rightarrow \int {{{\sec }^n}xdx} = {\sec ^{n - 2}}x\tan x - \left( {n - 2} \right)\int {{{\sec }^n}xdx} + \left( {n - 2} \right)\int {{{\sec }^{n - 2}}xdx}$
Move $\left( {n - 2} \right)\int {{{\sec }^n}xdx}$ to the left side and simplify it.
$\Rightarrow \left( {1 + n - 2} \right)\int {{{\sec }^n}xdx} = {\sec ^{n - 2}}x\tan x + \left( {n - 2} \right)\int {{{\sec }^{n - 2}}xdx}$
$\Rightarrow \int {{{\sec }^n}xdx} = \dfrac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}xdx}$
Second integral: $\int {\left( {{{\tan }^n}x} \right)dx}$
It can be written as $\int {{{\tan }^{n - 2}}x \cdot {{\tan }^2}xdx}$
Use identity ${\tan ^2}x + 1 = {\sec ^2}x$.
$\Rightarrow \int {{{\tan }^{n - 2}}x \cdot {{\tan }^2}xdx} = \int {{{\tan }^{n - 2}}x \cdot \left( {{{\sec }^2}x - 1} \right)dx}$
$\Rightarrow \int {{{\tan }^{n - 2}}x \cdot {{\tan }^2}xdx} = \int {{{\tan }^{n - 2}}x \cdot {{\sec }^2}xdx} - \int {{{\tan }^{n - 2}}xdx}$
Now, use integration by parts with $u = {\tan ^{n - 2}}x$ and $dv = {\sec ^2}xdx$…(vii)
Differentiate $u$ with respect to $x$.
$\dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}\left( {{{\tan }^{n - 2}}x} \right)$…(viii)
Now, using the differentiation formula $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$ in differentiation (viii), we get
$\dfrac{{du}}{{dx}} = \left( {n - 2} \right){\tan ^{n - 1}}x\left( {{{\sec }^2}x} \right)$
$\Rightarrow du = \left( {n - 2} \right){\tan ^{n - 1}}x{\sec ^2}x$…(ix)
Now, integrate $v$ with respect to $x$.
$\int {dv} = \int {{{\sec }^2}xdx}$…(x)
Now, using the integration formula $\int {{{\sec }^2}xdx} = \tan x$ in integral (x), we get
$v = \tan x$…(xi)
The integration by parts formula is:
$\int {udv} = uv - \int {vdu}$
Put the value of $u,v,du,dv$ from (vii), (ix) and (xi).
$\int {{{\tan }^{n - 2}}x \cdot {{\tan }^2}xdx} = {\tan ^{n - 2}}x\tan x - \int {\tan x\left( {n - 2} \right){{\tan }^{n - 1}}x{{\sec }^2}xdx} - \int {{{\tan }^{n - 2}}xdx}$
Using property that 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.
$\Rightarrow \int {{{\tan }^n}xdx} = {\tan ^{n - 1}}x - \left( {n - 2} \right)\int {{{\tan }^n}x{{\sec }^2}xdx} - \int {{{\tan }^{n - 2}}xdx}$…(xii)
Now, put the value of ${\sec ^2}x$ in integral (xii).
$\Rightarrow \int {{{\tan }^n}xdx} = {\tan ^{n - 1}}x - \left( {n - 2} \right)\int {{{\tan }^n}x\left( {{{\tan }^2}x + 1} \right)dx} - \int {{{\tan }^{n - 2}}xdx}$
$\Rightarrow \int {{{\tan }^n}xdx} = {\tan ^{n - 1}}x - \left( {n - 2} \right)\int {{{\tan }^{n + 2}}xdx - } \left( {n - 2} \right)\int {{{\tan }^n}xdx} - \int {{{\tan }^{n - 2}}xdx}$
Move $\left( {n - 2} \right)\int {{{\tan }^n}xdx}$ to the left side and simplify it.
$\Rightarrow \left( {1 + n - 2} \right)\int {{{\tan }^n}xdx} = {\tan ^{n - 1}}x + \left( {n - 1} \right)\int {{{\tan }^{n - 2}}xdx}$
$\Rightarrow \int {{{\tan }^n}xdx} = \dfrac{{{{\tan }^{n - 1}}x}}{{n - 1}} + \int {{{\tan }^{n - 2}}xdx}$
Final solution: Hence, $\int {{{\sec }^n}xdx} = \dfrac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \dfrac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}xdx}$ and $\int {{{\tan }^n}xdx} = \dfrac{{{{\tan }^{n - 1}}x}}{{n - 1}} + \int {{{\tan }^{n - 2}}xdx}$.

Note:
In the reduction formulas, we reduce the exponent of secant and tangent by $2$. Thus, if the formulas are applied repeatedly, the exponent can eventually be reduced to $0$ if $n$ is even or $1$ if $n$ is odd, at which point the integration can be completed.