Question

integration of cotx /sinx dx =​

Answer 1

Question :

Evaluate

$\sf\int\dfrac{\cot\:x}{\sin\:x}dx$

Answer:

We have to evaluate :

$\sf\int\dfrac{\cot\:x}{\sin\:x}dx$

$\sf\int\dfrac{\cos\:x}{\sin^2x}dx$

Let $\sf\:sin\:x=t$

Now, Differentiate it with respect to x

$\sf\implies\dfrac{dt}{dx}=\cos\:x$

$\sf\implies\:dt=\cos\:x\:dx$

Then ,

$\sf\int\dfrac{\cos\:x}{\sin^2x}dx$

$\sf=\int\dfrac{\cos\:x}{t^2}\times\dfrac{dt}{\cos\:x}$

$\sf=\int\dfrac{dt}{t^2}$

$\sf=\int\:t^{-2}dt$

We know that

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

Then ,

$\sf=\dfrac{t^{-2+1}}{-2+1}+c$

$\sf=-t^{-1}+c$

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

$\sf=\dfrac{-1}{\sin\:x}+c$

It is the required solution !

Answer 2

Answer:

The value of given integral is

$\int \frac{cotx}{sinx} = - \frac{1}{sinx} + c$

Explanation:

Given integral is

$I=\int \frac{cotx}{sinx} dx$

We know that,

$cotx = \frac{cosx}{sinx}$

Therefore,our integral becomes

$I=\int \frac{cosx}{sin {}^{2}x } dx$

We solve this by using substitution method.Let us assume that

$sinx = z \\ = > cosx.dx = dz$

Substituting these terms in above integral,we get

$I=\int \frac{dz}{z {}^{2} } \\ = \int {z}^{ - 2} dz \\ = \frac{ {z}^{ - 2 + 1} }{ - 2 + 1} + c = - \frac{1}{z} + c$

Substituting the value of z we get the integral as

$- \frac{1}{sinx} + c$

Therefore,the value of given integral is

$\int \frac{cotx}{sinx} = - \frac{1}{sinx} + c$

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