Answer

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Hint: Convert the given integral to ‘sin’ and ‘cos’ functions, now divide numerator and denominator by ${{\cos }^{4}}x$ . Later solve the integral to get the answer.

Complete step-by-step answer:

Here, integral given is

$\int{{{\sec }^{2}}x{{\csc }^{2}}xdx}$

Let given integral be I, hence

$I=\int{{{\sec }^{2}}x{{\csc }^{2}}xdx}$……………(i)

We know,

$\sec x=\dfrac{1}{\cos x}$ and $cscx=\dfrac{1}{\sin x}$

Hence, I can be written as

$I=\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

Now, let us convert the given integral to $\sec x$ and $\tan x$ by dividing the numerator and denominator \[{{\cos }^{4}}x\]and using the relation $\tan x=\dfrac{\sin x}{\cos x}$ .

Hence, we get

$I=\int{\dfrac{\dfrac{1}{{{\cos }^{4}}x}}{\dfrac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{\cos }^{4}}x}}}dx=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}}}dx$

$I=\int{\dfrac{{{\sec }^{4}}x}{{{\tan }^{2}}x}}dx$

Or

$I=\int{\dfrac{{{\sec }^{2}}x{{\sec }^{2}}x}{{{\tan }^{2}}x}}dx$

Now, we can convert one \['{{\sec }^{2}}x'\] to \['1+{{\tan }^{2}}x'\] in numerator by using trigonometric identity given, so the above expression can be written as,

\[I=\int{\dfrac{(1+ta{{n}^{2}}x)se{{c}^{2}}x}{{{\tan }^{2}}x}}dx\]…………………. (ii)

Now, we can suppose ‘tan x’ as ‘t’ and get derivative of it as ‘${{\sec }^{2}}x$’ so that

We can get an integral in ‘x’ only.

Let $t=\tan x$

Differentiating both sides w.r.t ‘x’, we get

As we know $\dfrac{d}{dx}(tanx)=se{{c}^{2}}x$, the above expression can be written as,

$\dfrac{dt}{dx}=se{{c}^{2}}x$

Now, cross multiplying above equation, we get

$dt={{\sec }^{2}}xdx$

Hence, integral ‘I’ can be re-written in terms of ‘t’ as

$I=\int{\dfrac{1+{{t}^{2}}}{{{t}^{2}}}}dt=\int{\left( \dfrac{1}{{{t}^{2}}}+\dfrac{{{t}^{2}}}{{{t}^{2}}} \right)}dt$

or

$I=\int{\left( \dfrac{1}{{{t}^{2}}}+1 \right)}dt$

Now we know that

$\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$

Hence, integral can be simplified as

$I=\int{{{t}^{-{{2}^{{}}}}}dt+\int{{{t}^{o}}dt}}$

Using the above identity of integral, we get

$I=\dfrac{{{t}^{-2+1}}}{-2+1}+\dfrac{{{t}^{o+1}}}{0+1}+c$

$I=\dfrac{{{t}^{-1}}}{-1}+\dfrac{{{t}^{1}}}{1}+c$

Or

$I=\dfrac{-1}{t}+t+c$

Now, putting value of t as $\tan x$ in the above equation, we can get value of I as

$I=\dfrac{-1}{\operatorname{tanx}}+\tan x+c$

Hence,

$\int{{{\sec }^{2}}{{\csc }^{2}}xdx=\dfrac{-1}{\tan x}+\tan x+c}$

Note: Diving numerator and denominator by${{\cos }^{4}}x$ of integral $\int{\dfrac{1}{{{\sin }^{2}}xco{{s}^{2}}x}}$ is the key point of the solution. Try to convert these kind of questions to $\tan x$ and ${{\sec }^{2}}x$ if power of $\sin x$ and $\cos x$ are even so that we can always substitute $\tan x$ and get derivative as ${{\sec }^{2}}x$.

Another approach for the given integration would be that we can convert ${{\sec }^{2}}x$ to $1+{{\tan }^{2}}x$, then $\tan x$ to $\dfrac{1}{\cot x}$, Now, suppose $\cot x=t$ expression would be

$\int{\left( 1+\dfrac{1}{{{\cot }^{2}}x} \right)}{{\csc }^{2}}xdx$ and solve accordingly.

Complete step-by-step answer:

Here, integral given is

$\int{{{\sec }^{2}}x{{\csc }^{2}}xdx}$

Let given integral be I, hence

$I=\int{{{\sec }^{2}}x{{\csc }^{2}}xdx}$……………(i)

We know,

$\sec x=\dfrac{1}{\cos x}$ and $cscx=\dfrac{1}{\sin x}$

Hence, I can be written as

$I=\int{\dfrac{1}{{{\sin }^{2}}x{{\cos }^{2}}x}dx}$

Now, let us convert the given integral to $\sec x$ and $\tan x$ by dividing the numerator and denominator \[{{\cos }^{4}}x\]and using the relation $\tan x=\dfrac{\sin x}{\cos x}$ .

Hence, we get

$I=\int{\dfrac{\dfrac{1}{{{\cos }^{4}}x}}{\dfrac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{\cos }^{4}}x}}}dx=\int{\dfrac{{{\sec }^{4}}x}{\dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}\dfrac{{{\cos }^{2}}x}{{{\cos }^{2}}x}}}dx$

$I=\int{\dfrac{{{\sec }^{4}}x}{{{\tan }^{2}}x}}dx$

Or

$I=\int{\dfrac{{{\sec }^{2}}x{{\sec }^{2}}x}{{{\tan }^{2}}x}}dx$

Now, we can convert one \['{{\sec }^{2}}x'\] to \['1+{{\tan }^{2}}x'\] in numerator by using trigonometric identity given, so the above expression can be written as,

\[I=\int{\dfrac{(1+ta{{n}^{2}}x)se{{c}^{2}}x}{{{\tan }^{2}}x}}dx\]…………………. (ii)

Now, we can suppose ‘tan x’ as ‘t’ and get derivative of it as ‘${{\sec }^{2}}x$’ so that

We can get an integral in ‘x’ only.

Let $t=\tan x$

Differentiating both sides w.r.t ‘x’, we get

As we know $\dfrac{d}{dx}(tanx)=se{{c}^{2}}x$, the above expression can be written as,

$\dfrac{dt}{dx}=se{{c}^{2}}x$

Now, cross multiplying above equation, we get

$dt={{\sec }^{2}}xdx$

Hence, integral ‘I’ can be re-written in terms of ‘t’ as

$I=\int{\dfrac{1+{{t}^{2}}}{{{t}^{2}}}}dt=\int{\left( \dfrac{1}{{{t}^{2}}}+\dfrac{{{t}^{2}}}{{{t}^{2}}} \right)}dt$

or

$I=\int{\left( \dfrac{1}{{{t}^{2}}}+1 \right)}dt$

Now we know that

$\int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}}$

Hence, integral can be simplified as

$I=\int{{{t}^{-{{2}^{{}}}}}dt+\int{{{t}^{o}}dt}}$

Using the above identity of integral, we get

$I=\dfrac{{{t}^{-2+1}}}{-2+1}+\dfrac{{{t}^{o+1}}}{0+1}+c$

$I=\dfrac{{{t}^{-1}}}{-1}+\dfrac{{{t}^{1}}}{1}+c$

Or

$I=\dfrac{-1}{t}+t+c$

Now, putting value of t as $\tan x$ in the above equation, we can get value of I as

$I=\dfrac{-1}{\operatorname{tanx}}+\tan x+c$

Hence,

$\int{{{\sec }^{2}}{{\csc }^{2}}xdx=\dfrac{-1}{\tan x}+\tan x+c}$

Note: Diving numerator and denominator by${{\cos }^{4}}x$ of integral $\int{\dfrac{1}{{{\sin }^{2}}xco{{s}^{2}}x}}$ is the key point of the solution. Try to convert these kind of questions to $\tan x$ and ${{\sec }^{2}}x$ if power of $\sin x$ and $\cos x$ are even so that we can always substitute $\tan x$ and get derivative as ${{\sec }^{2}}x$.

Another approach for the given integration would be that we can convert ${{\sec }^{2}}x$ to $1+{{\tan }^{2}}x$, then $\tan x$ to $\dfrac{1}{\cot x}$, Now, suppose $\cot x=t$ expression would be

$\int{\left( 1+\dfrac{1}{{{\cot }^{2}}x} \right)}{{\csc }^{2}}xdx$ and solve accordingly.

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