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,
\[\begin{align}
  & I=\int{\left( \dfrac{1}{{{\cos }^{2}}x} \right)}.\left( \dfrac{{{\sin }^{2}}x}{1} \right) \\
 & \\
 & I=\int{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}}.dx\text{ }~\to (2) \\
\end{align}\]

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
\[\begin{align}
  & {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\
 & \\
 & \therefore {{\tan }^{2}}x={{\sec }^{2}}x-1 \\
\end{align}\]

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 \[{{\sec }^{2}}x\].