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Hint: This is a standard integral and expected to be memorized. Try substituting $t=\tan \dfrac{x}{2}$ and use the fact that $\csc x=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2\tan \dfrac{x}{2}}$. Alternatively, multiply the numerator and denominator by cosec x - cot x and put t = cosec x - cot x.

Complete step-by-step answer:

Let $t=\tan \dfrac{x}{2}$

Differentiating both sides, we get

$dt=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx$

We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

Using the above formula, we get

\[\begin{align}

& dt=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2}=\dfrac{1+{{t}^{2}}}{2}dx \\

& \Rightarrow dx=\dfrac{2dt}{1+{{t}^{2}}} \\

\end{align}\]

Also $\csc x=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2\tan \dfrac{x}{2}}=\dfrac{1+{{t}^{2}}}{2t}$

Hence we have

$\begin{align}

& \int{\csc xdx}=\int{\dfrac{2dt}{1+{{t}^{2}}}\times }\dfrac{1+{{t}^{2}}}{2t} \\

& =\int{\dfrac{dt}{t}} \\

\end{align}$

We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$

Using we get

$\int{\csc xdx}=\ln \left| t \right|+C$

Reverting to the original variable, we get

$\int{\csc xdx}=\ln \left| \tan \dfrac{x}{2} \right|+C$

Note: [1] Alternatively we can solve the above question multiplying the numerator and denominator by cosec x – cot x and substituting cosec x – cot x = t

We have $\csc x=\dfrac{\csc x\left( \csc x-\cot x \right)}{\csc x-\cot x}$

Put cosec x - cot x = t

Differentiating both sides we get

$\left( -\csc x\cot x+{{\csc }^{2}}x \right)dx=dt$

Taking cosec x common, we get

$\Rightarrow \csc x\left( \csc x-\cot x \right)dx=dt$

Hence we have

$\int{\csc xdx=\int{\dfrac{dt}{t}}}$

We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$

Using we get

$\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+C$

[2] Although the two results look completely different from each other, by Lagrange mean value theorem, they differ only by a constant.

Hence $\ln \left| \csc x-\cot x \right|=\ln \left| \tan \dfrac{x}{2} \right|+C$ for some constant C.

i.e. $\csc x-\cot x=A\tan \dfrac{x}{2}$ for some constant A. It can be verified that A = 1.

[3] Proof of [2]: Let F(x) and G(x) be two antiderivatives of the function f(x).

Applying Lagrange mean value theorem to the function F(x) – G(x) in the interval [0,y]{Assuming continuity at 0}, we get

\[\dfrac{F(y)-G(y)-F(0)+G(0)}{y}=f(c)-f(c)=0\]

i.e. $F(y)-G(y)-F(0)+G(0)=0$

Since F(0)-G(0) is a constant let C = F(0) – G(0), we get

$F(y)=G(y)+C$ or in other words F(x) and G(x) differ only by a constant.

[4] Some fundamental integrals to remember

[a] $\int{\sin x}dx=-\cos x+C$

[b] $\int{\cos xdx}=\sin x+C$

[c] $\int{\tan xdx}=\ln \sec x+C$

[d] \[\int{\cot x}dx=\ln \sin x+C\]

[e] \[\int{\sec xdx}=\ln \left| \sec x+\tan x \right|+C\]

[f] $\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+C$

[g] $\int{{{\sec }^{2}}x}dx=\tan x+C$

[h] $\int{{{\csc }^{2}}xdx}=-\cot x+C$

[i] $\int{\dfrac{dx}{\sqrt{1-{{x}^{2}}}}=\arcsin x+C}$

[j] $\int{\dfrac{dx}{{{x}^{2}}+1}=\arctan x+C}$

Complete step-by-step answer:

Let $t=\tan \dfrac{x}{2}$

Differentiating both sides, we get

$dt=\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx$

We know that ${{\sec }^{2}}x=1+{{\tan }^{2}}x$

Using the above formula, we get

\[\begin{align}

& dt=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2}=\dfrac{1+{{t}^{2}}}{2}dx \\

& \Rightarrow dx=\dfrac{2dt}{1+{{t}^{2}}} \\

\end{align}\]

Also $\csc x=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2\tan \dfrac{x}{2}}=\dfrac{1+{{t}^{2}}}{2t}$

Hence we have

$\begin{align}

& \int{\csc xdx}=\int{\dfrac{2dt}{1+{{t}^{2}}}\times }\dfrac{1+{{t}^{2}}}{2t} \\

& =\int{\dfrac{dt}{t}} \\

\end{align}$

We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$

Using we get

$\int{\csc xdx}=\ln \left| t \right|+C$

Reverting to the original variable, we get

$\int{\csc xdx}=\ln \left| \tan \dfrac{x}{2} \right|+C$

Note: [1] Alternatively we can solve the above question multiplying the numerator and denominator by cosec x – cot x and substituting cosec x – cot x = t

We have $\csc x=\dfrac{\csc x\left( \csc x-\cot x \right)}{\csc x-\cot x}$

Put cosec x - cot x = t

Differentiating both sides we get

$\left( -\csc x\cot x+{{\csc }^{2}}x \right)dx=dt$

Taking cosec x common, we get

$\Rightarrow \csc x\left( \csc x-\cot x \right)dx=dt$

Hence we have

$\int{\csc xdx=\int{\dfrac{dt}{t}}}$

We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$

Using we get

$\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+C$

[2] Although the two results look completely different from each other, by Lagrange mean value theorem, they differ only by a constant.

Hence $\ln \left| \csc x-\cot x \right|=\ln \left| \tan \dfrac{x}{2} \right|+C$ for some constant C.

i.e. $\csc x-\cot x=A\tan \dfrac{x}{2}$ for some constant A. It can be verified that A = 1.

[3] Proof of [2]: Let F(x) and G(x) be two antiderivatives of the function f(x).

Applying Lagrange mean value theorem to the function F(x) – G(x) in the interval [0,y]{Assuming continuity at 0}, we get

\[\dfrac{F(y)-G(y)-F(0)+G(0)}{y}=f(c)-f(c)=0\]

i.e. $F(y)-G(y)-F(0)+G(0)=0$

Since F(0)-G(0) is a constant let C = F(0) – G(0), we get

$F(y)=G(y)+C$ or in other words F(x) and G(x) differ only by a constant.

[4] Some fundamental integrals to remember

[a] $\int{\sin x}dx=-\cos x+C$

[b] $\int{\cos xdx}=\sin x+C$

[c] $\int{\tan xdx}=\ln \sec x+C$

[d] \[\int{\cot x}dx=\ln \sin x+C\]

[e] \[\int{\sec xdx}=\ln \left| \sec x+\tan x \right|+C\]

[f] $\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+C$

[g] $\int{{{\sec }^{2}}x}dx=\tan x+C$

[h] $\int{{{\csc }^{2}}xdx}=-\cot x+C$

[i] $\int{\dfrac{dx}{\sqrt{1-{{x}^{2}}}}=\arcsin x+C}$

[j] $\int{\dfrac{dx}{{{x}^{2}}+1}=\arctan x+C}$

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