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mr_godi [17]
3 years ago
5

Please help me out quickly!​

Mathematics
1 answer:
VLD [36.1K]3 years ago
7 0

Answer:

imvnuifdv

Step-by-step explanation:

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∫(cosx) / (sin²x) dx
kirza4 [7]
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2822772

_______________


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)}


Make the following substitution:

\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}


and then, the integral (i) becomes

=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}


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}


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}


\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark


I hope this helps. =)


Tags:  <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

5 0
3 years ago
1/5+1/4+2/5 this is adding fractions with unlike denominators what is the answer
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What you would do is change all the fractions to 20ths so they have a common denominator. For 1/5, you multiply 4x5 to get 20 in the denominator. Whatever you multiply on the bottom, you must do at the top. Do that for all fractions and you would get this:
4/20 + 5/20 + 8/20= 17/20 because denominator stays the same, add the values in the numerator.
ANSWER: 17/20
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3 years ago
7a-3b-a+8-9b<br> Pls help!
iVinArrow [24]

Answer:

6a-12b+8

Step-by-step explanation:

Im pretty sure this is the answer but if it's not im so sorry

7 0
2 years ago
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