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Dovator [93]
3 years ago
5

Simplify 5+2i/6+i

Mathematics
1 answer:
nlexa [21]3 years ago
3 0

Given problem is \frac{(5+2i)}{(6+i)}

To simplify that we need to multiply and divide by conjugate of the denominator

conjugate of denominator 6+i will be 6-i as we just need to change sign of the imaginary part

Now multiply and divide by 6-i


=\frac{(5+2i)}{(6+i)}\cdot\frac{(6-i)}{(6-i)}

=\frac{30-5i+12i-2i^2}{36-6i+6i-i^2}

=\frac{30+7i-2i^2}{36-i^2}

=\frac{30+7i-2\left(-1\right)}{36-\left(-1\right)}

=\frac{30+7i+2}{36+1}

=\frac{32+7i}{37}

=\frac{32}{37}+\frac{7}{37}i

Hence final answer is \frac{32}{37}+\frac{7}{37}i

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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}\\\\\\&#10;=\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}\\\\\\&#10;=\mathsf{\displaystyle\int\! u^{-2}\,du}


Integrate it by applying the power rule:

\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\&#10;\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\&#10;\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}\\\\\\&#10;\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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