Question

Evaluate the given integral
$\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}}.dx$

Answer 1

Hint: Apply basic trigonometric formula and make the given expression in terms of ${{\tan }^{2}}x$. Thus use the trigonometric formula of ${{\tan }^{2}}x$ and find the integral of the expression obtained.

Complete step-by-step answer:

Given to us the integral, which we need to evaluate. Let us take the integral as equal to $I$.
$\therefore I=\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}}.dx$ $\to$ (1)
We know that $\sec x=\dfrac{1}{\cos x}$.
Similarly $\operatorname{cosec}x=\dfrac{1}{\sin x}$.
Now apply the values of $\sec x$ and $\operatorname{cosec}x$ in equation (1)
This $I$ change to,
$I=\int{\dfrac{\dfrac{1}{{{\cos }^{2}}x}}{\dfrac{1}{\sin {{x}^{2}}}}}.dx$
Now this is of the form$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}$. We can rewrite it as,
$\dfrac{a}{b}\times \dfrac{d}{c}$, where $a=1,b={{\cos }^{2}}x,c=1,d={{\sin }^{2}}x$.
Thus we can write it as,

& I=\int{\left( \dfrac{1}{{{\cos }^{2}}x} \right)}.\left( \dfrac{{{\sin }^{2}}x}{1} \right) \\\ & \\\ & I=\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}.dx\text{ }~\to (2) \\\ \end{aligned}$$ We know the basic trigonometric formulas that is $$\dfrac{\sin x}{\cos x}=\tan x$$. Thus equation (2) change to $$I=\int{{{\tan }^{2}}x.dx}$$ We know that $$\begin{aligned} & {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\\ & \\\ & \therefore {{\tan }^{2}}x={{\sec }^{2}}x-1 \\\ \end{aligned}$$ Thus put $${{\tan }^{2}}x={{\sec }^{2}}x-1$$ $$\therefore I=\int{\left( {{\sec }^{2}}x-1 \right)dx}$$ $$=\int{{{\sec }^{2}}x.dx}-\int{1.dx=\int{{{\sec }^{2}}x.dx}-\int{{{x}^{0}}}}.dx$$ $$\to (3)$$ We know that $$\int{{{\sec }^{2}}x.dx=\tan x}$$ We know that $${{x}^{0}}=1$$ Similarly $$\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}+c$$ $$I=\int{{{\sec }^{2}}x.dx-\int{{{x}^{0}}.dx}}$$ $$=\tan x-\dfrac{{{x}^{0+1}}}{0+1}+c$$ $$=\tan x-\dfrac{{{x}^{1}}}{1}+c$$ $$I=\tan x-x+c$$ Thus we got$$\int{\dfrac{{{\sec }^{2}}x}{\cos e{{c}^{2}}x}}.dx=\tan x-x+c$$. Hence we have solved the integral. Note: We have basic trigonometric formulas to solve this particular question. You should remember them or else you won’t be able to solve them. Here we took $${{\tan }^{2}}x={{\sec }^{2}}x-1$$ in order to find the integral of $${{\tan }^{2}}x$$. There is no direct integration for $${{\tan }^{2}}x$$, thus converting in terms of $${{\sec }^{2}}x$$.