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Evaluate the indefinite integral:
![\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\ =\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20%5Cfrac%7Bcos%5C%2Cx%7D%7Bsin%5E2%5C%2Cx%7D%5C%2Cdx%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20%5Cfrac%7B1%7D%7B%28sin%5C%2Cx%29%5E2%7D%5Ccdot%20cos%5C%2Cx%5C%2Cdx%5Cqquad%5Cquad%28i%29%7D)
Make the following substitution:
![\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}](https://tex.z-dn.net/?f=%5Cmathsf%7Bsin%5C%2Cx%3Du%5Cquad%5CRightarrow%5Cquad%20cos%5C%2Cx%5C%2Cdx%3Ddu%7D)
and then, the integral (i) becomes
![=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\ =\mathsf{\displaystyle\int\! u^{-2}\,du}](https://tex.z-dn.net/?f=%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20%5Cfrac%7B1%7D%7Bu%5E2%7D%5C%2Cdu%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20u%5E%7B-2%7D%5C%2Cdu%7D)
Integrate it by applying the power rule:
![\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\ \mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\ \mathsf{=-\,\dfrac{1}{u}+C}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3D%5Cdfrac%7Bu%5E%7B-2%2B1%7D%7D%7B-2%2B1%7D%2BC%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D%5Cdfrac%7Bu%5E%7B-1%7D%7D%7B-1%7D%2BC%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D-%5C%2C%5Cdfrac%7B1%7D%7Bu%7D%2BC%7D)
Now, substitute back for u = sin x, so the result is given in terms of x:
![\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\ \mathsf{=-\,csc\,x+C}](https://tex.z-dn.net/?f=%5Cmathsf%7B%3D-%5C%2C%5Cdfrac%7B1%7D%7Bsin%5C%2Cx%7D%2BC%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%3D-%5C%2Ccsc%5C%2Cx%2BC%7D)
![\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Ctherefore~~%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdisplaystyle%5Cint%5C%21%20%5Cfrac%7Bcos%5C%2Cx%7D%7Bsin%5E2%5C%2Cx%7D%5C%2Cdx%3D-%5C%2Ccsc%5C%2Cx%2BC%7D%20%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
I hope this helps. =)
Tags: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>
The first one is 5.3 you multiply 4x4 which is 16 then divide by 3 which then gives you the answer
Answer:
y = 6
Step-by-step explanation:
If the slope of a line is zero, then it is a horizontal line. This means that the line will have a constant y value for all x, hence the equation of the line is y = 6
First equation answer:
(j= -7)
Second equation answer:
(j= 5)