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Last updated date: 02nd Dec 2023
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# Find the integral of the following function$\csc x$

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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.

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