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Hint: If you can not integrate a term then break this term into two or more terms then it will be easy to integrate.

We have to find $\int {\cot x dx} $

We can write :$\cot x = \dfrac{{\cos x}}{{\sin x}}$

$\int {\dfrac{{\cos x}}{{\sin x}}dx} $$ \ldots \ldots \left( 1 \right)$

Now, to make it easy we will use substitution method

Put,$\sin x = t$

Differentiate it with respect to x.

$ \Rightarrow \cos x dx = dt$

Now after putting this value in equation $\left( 1 \right)$ we get,

$\int {\dfrac{1}{t}dt = \log t + c = \log \left| {\sin x} \right| + c} $

Here we use modulus, because if you want to take log of

Any number, you have to make sure the number is positive .

Note:

If your question becomes a fractional question then it’s higher chances to be solved by substitution method and then use substitution to solve the integration and again substitute variables with original terms in which questions are given.

We have to find $\int {\cot x dx} $

We can write :$\cot x = \dfrac{{\cos x}}{{\sin x}}$

$\int {\dfrac{{\cos x}}{{\sin x}}dx} $$ \ldots \ldots \left( 1 \right)$

Now, to make it easy we will use substitution method

Put,$\sin x = t$

Differentiate it with respect to x.

$ \Rightarrow \cos x dx = dt$

Now after putting this value in equation $\left( 1 \right)$ we get,

$\int {\dfrac{1}{t}dt = \log t + c = \log \left| {\sin x} \right| + c} $

Here we use modulus, because if you want to take log of

Any number, you have to make sure the number is positive .

Note:

If your question becomes a fractional question then it’s higher chances to be solved by substitution method and then use substitution to solve the integration and again substitute variables with original terms in which questions are given.

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