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Question

Question: What is the value of integral \(\int{\cot x\cos xdx}?\)...

What is the value of integral cotxcosxdx?\int{\cot x\cos xdx}?

Explanation

Solution

We have already learnt the trigonometric identity given by cotx=cosxsinx.\cot x=\dfrac{\cos x}{\sin x}. And we also know the trigonometric identity given by sin2x+cos2x=1.{{\sin }^{2}}x+{{\cos }^{2}}x=1. We will rearrange the identities if necessary. Then we will use the linearity property of integration.

Complete step-by-step solution:
Let us consider the given integral cotxcosxdx.\int{\cot x\cos xdx}.
In order to find the integral, we need to simplify the function under the integral side.
So, we will use the trigonometric identity given by cotx=cosxsinx.\cot x=\dfrac{\cos x}{\sin x}.
Let us substitute this in the given integral.
Then, we will get cotxcosxdx=cosxsinxcosxdx.\int{\cot x\cos xdx}=\int{\dfrac{\cos x}{\sin x}\cos xdx}.
Now, we can change this into cotxcosxdx=cos2xsinxdx.\int{\cot x\cos xdx}=\int{\dfrac{{{\cos }^{2}}x}{\sin x}dx}.
We have already learnt the Pythagorean identity given by sin2x+cos2x=1.{{\sin }^{2}}x+{{\cos }^{2}}x=1.
Now, if we rearrange this trigonometric identity, we will get cos2x=1sin2x.{{\cos }^{2}}x=1-{{\sin }^{2}}x.
Let us substitute this identity in the given integral.
Then, we will get cotxcosxdx=1sin2xsinxdx.\int{\cot x\cos xdx}=\int{\dfrac{1-{{\sin }^{2}}x}{\sin x}dx}.
From this, we will get cotxcosxdx=(1sinxsin2xsinx)dx.\int{\cot x\cos xdx}=\int{\left( \dfrac{1}{\sin x}-\dfrac{{{\sin }^{2}}x}{\sin x} \right)dx}.
Now let us use the linearity property of the integrals.
We know that the linear property is given by (f+g)dx=fdx+gdx\int{\left( f+g \right)dx}=\int{fdx}+\int{gdx} where ff and gg are functions of x.x.
Now, when we apply this linearity property, we will get cotxcosxdx=1sinxdxsin2xsinxdx.\int{\cot x\cos xdx}=\int{\dfrac{1}{\sin x}dx}-\int{\dfrac{{{\sin }^{2}}x}{\sin x}dx}.
We know that 1sinx=cosecx\dfrac{1}{\sin x}=\cos ecx and sin2xsinx=sinx.\dfrac{{{\sin }^{2}}x}{\sin x}=\sin x.
Now, when we substitute these identities, we will get cotxcosxdx=cosecxdxsinxdx.\int{\cot x\cos xdx}=\int{\cos ecxdx}-\int{\sin xdx}.
Let us recall the integrals given by sinxdx=cosx+C\int{\sin xdx}=-\cos x+C and cosecxdx=lncosecx+cotx+C\int{\cos ecx}dx=-\ln \left| \cos ecx+\cot x \right|+C
Now, we are going to apply the above written identities in the obtained integral.
Then, as a result, we will get cotxcosxdx=lncosecx+cotx(cosx)+C\int{\cot x\cos xdx}=-\ln \left| \cos ecx+\cot x \right|-\left( -\cos x \right)+C where CC is the constant of integration.
And this will give us cotxcosxdx=lncosecx+cotx+cosx+C.\int{\cot x\cos xdx}=-\ln \left| \cos ecx+\cot x \right|+\cos x+C.
Hence the required integral is cotxcosxdx=lncosecx+cotx+cosx+C.\int{\cot x\cos xdx}=-\ln \left| \cos ecx+\cot x \right|+\cos x+C.

Note: We should not forget to put the constant of integration when we deal with the indefinite integrals where there are no upper limit and lower limit given. The integrals with upper limit and lower limit are called the definite integrals.