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olga nikolaevna [1]
3 years ago
5

Correct answer gets 5 star and thanks.

Mathematics
1 answer:
masha68 [24]3 years ago
8 0
15/p;
4(px+1) / 4 = 64 / 4
= px+1 = 16
-1 -1
= px = 15; p ≠ 0
/p /p
= x = 15/p; p ≠ 0
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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>

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Yes it is d for sure and glad to see you on brainly!
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