Question

Find the integral of the following function
$\csc x$

Answer 1

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

& dt=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2}=\dfrac{1+{{t}^{2}}}{2}dx \\\ & \Rightarrow dx=\dfrac{2dt}{1+{{t}^{2}}} \\\ \end{aligned}$$ Also $\csc x=\dfrac{1+{{\tan }^{2}}\dfrac{x}{2}}{2\tan \dfrac{x}{2}}=\dfrac{1+{{t}^{2}}}{2t}$ Hence we have $\begin{aligned} & \int{\csc xdx}=\int{\dfrac{2dt}{1+{{t}^{2}}}\times }\dfrac{1+{{t}^{2}}}{2t} \\\ & =\int{\dfrac{dt}{t}} \\\ \end{aligned}$ We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$ Using we get $\int{\csc xdx}=\ln \left| t \right|+C$ Reverting to the original variable, we get $\int{\csc xdx}=\ln \left| \tan \dfrac{x}{2} \right|+C$ Note: [1] Alternatively we can solve the above question multiplying the numerator and denominator by cosec x – cot x and substituting cosec x – cot x = t We have $\csc x=\dfrac{\csc x\left( \csc x-\cot x \right)}{\csc x-\cot x}$ Put cosec x - cot x = t Differentiating both sides we get $\left( -\csc x\cot x+{{\csc }^{2}}x \right)dx=dt$ Taking cosec x common, we get $\Rightarrow \csc x\left( \csc x-\cot x \right)dx=dt$ Hence we have $\int{\csc xdx=\int{\dfrac{dt}{t}}}$ We know that $\int{\dfrac{dx}{x}}=\ln \left| x \right|+C$ Using we get $\int{\csc xdx}=\ln \left| \csc x-\cot x \right|+C$ [2] Although the two results look completely different from each other, by Lagrange mean value theorem, they differ only by a constant. Hence $\ln \left| \csc x-\cot x \right|=\ln \left| \tan \dfrac{x}{2} \right|+C$ for some constant C. i.e. $\csc x-\cot x=A\tan \dfrac{x}{2}$ for some constant A. It can be verified that A = 1. [3] Proof of [2]: Let F(x) and G(x) be two antiderivatives of the function f(x). Applying Lagrange mean value theorem to the function F(x) – G(x) in the interval [0,y]{Assuming continuity at 0}, we get $$\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] $\int{\dfrac{dx}{{{x}^{2}}+1}=\arctan x+C}$