Question

Evaluate the given integral $\int{\left( {{\sec }^{n}}x.\tan x \right)dx}$

Answer 1

Hint: For solving questions of this type we should have the integrations and derivatives basic formulas in our mind otherwise we will face difficulties. Here in this question we will assume $\sec x=t$ and use the following two formulae of differentiation and integrations as $\dfrac{d}{dx}\sec x=\sec x.\tan x$ and $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$ for $n\ne 1$. And perform further simplifications in order to obtain the answer.

Complete step-by-step solution
Let us assume that $\sec x=t$
By differentiating this on both sides we will get
$\sec x\tan xdx=dt$
Since we know that$\dfrac{d}{dx}\sec x=\sec x.\tan x$
So we can simply write this as
$\int{\left( {{\sec }^{n}}x.\tan x \right)dx=}\int{{{\sec }^{n-1}}x\left( \sec x\tan x \right)dx}$
This can be further simply written by using the assumption we made$\sec x=t$as
$\int{{{\sec }^{n-1}}x\left( \sec x\tan x \right)dx}=\int{{{t}^{n-1}}.dt}$
As we know the formula $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$ for $n\ne -1$
By using this $\int{{{\sec }^{n-1}}x\left( \sec x\tan x \right)dx}=\int{{{t}^{n-1}}.dt}$ this can be further simplified as
$\int{{{t}^{n-1}}.dt}=\dfrac{{{t}^{n}}}{n}+C$
By again substituting the value of $\sec x=t$ that we assumed before we will get
$\dfrac{{{t}^{n}}}{n}+C=\dfrac{{{\sec }^{n}}x}{n}+C$
Hence we can conclude that $\int{\left( {{\sec }^{n}}x.\tan x \right)dx}=\dfrac{{{\sec }^{n}}x}{n}+C$

Note: While solving this question we should note that $\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+C$ for $n\ne -1$
For $n=-1$we have a different formulae that is$\int{{{x}^{-1}}}dx=\log x+C$. And let us discuss one another point that the letter “C” which we are using often is a constant, this is a tricky point many times forgotten. Let us discuss some more formulae of trigonometric differential and integration functions that are of sine and cosine and secant and cosecant and tangent and cotangent functions of angles, their integrations and their differ nations respectively.
$\dfrac{d}{dx}\sin x=\cos x\And \dfrac{d}{dx}\cos x=-\sin x\And \dfrac{d}{dx}\tan x={{\sec }^{2}}x\And \dfrac{d}{dx}\cot x=-\cos e{{c}^{2}}x$
and $\dfrac{d}{dx}\cos ecx=-\cos ecx\cot x\And \dfrac{d}{dx}\sec x=\sec x\tan x$
Formulas’ for integration are: $\int{\sin xdx}=-\cos x+C\And \int{\cos x}dx=\sin x+C\And \int{\tan x}dx=-\log \left( \cos x \right)+C$
And $\int{\cos ecxdx}=-\log \left( \cos ecx-\cot x \right)+C\And \int{\sec x}dx=\log \left( \sec x+\tan x \right)+C\And \int{\cot x}dx=\log \left( \sin x \right)+C$Let us recall some more trigonometric basic formulae they are ${{\sin }^{2}}x+{{\cos }^{2}}x=1$and $\cos ecx=\dfrac{1}{\sin x}$ and $\sec x=\dfrac{1}{\cos x}$and $\tan x=\dfrac{\sin x}{\cos x}$.