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Serga [27]
3 years ago
6

How do I factor this polynomial? q^2-121

Mathematics
1 answer:
Tresset [83]3 years ago
5 0
q^2-121 =q^2-11^2=(q-11)(q+11)\\ \\ \\a^2-b^2 =(a-b)(a+b)


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Diego has 3 times as many comic books as Han.
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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
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