What is wrong in my answer? Subject: finding the integral of cot⁡x∫cot⁡xdx=∫cos⁡xsin⁡xsin2xdxAssume t=sin2x. Then dt=2sin⁡xcos⁡xdx. Using the identity:sin⁡2x=2sin⁡xcos⁡x∫cot⁡xdx=∫12tdt=12ln⁡(sin⁡x)2+c

Note that12ln⁡(sin⁡(x))2+c=ln|sin⁡(x)|+c

You can do even less: notice that cot⁡x=cos⁡xsin⁡x and that the derivative of sin⁡ is cos⁡. Hence∫cot⁡xdx=∫cos⁡xsin⁡xdx=log|sin⁡x|+c,c∈R

Note that12ln⁡(sin⁡(x))2+c=ln|sin⁡(x)|+c

You can do even less: notice that cot⁡x=cos⁡xsin⁡x and that the derivative of sin⁡ is cos⁡. Hence∫cot⁡xdx=∫cos⁡xsin⁡xdx=log|sin⁡x|+c,c∈R

"What is wrong in my answer? Subject: finding..." was answered by students like you

Ezra Phillips

2022-03-14

What is wrong in my answer? Subject: finding the integral of

\cot x

\int \cot x d x = \int \frac{\cos x \sin x}{\sin^{2} x} d x

Assume

t = \sin^{2} x

. Then

d t = 2 \sin x \cos x d x

. Using the identity:

\sin 2 x = 2 \sin x \cos x

\int \cot x d x = \int \frac{1}{2 t} d t = \frac{1}{2} {\ln (\sin x)}^{2} + c

alifutlessect692

Beginner2022-03-15Added 9 answers

Note that

{\frac{12}{\ln (\sin (x))}}^{2} + c = \ln | \sin (x) | + c

produsint8l5

Beginner2022-03-16Added 4 answers

You can do even less: notice that

\cot x = \frac{\cos x}{\sin x}

and that the derivative of

\sin is \cos

. Hence

\int \cot x d x = \int \frac{\cos x}{\sin x} d x = \log | \sin x | + c, c \in R

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