Question

∫(cosx) / (sin²x) dx

Answer 1

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\ =\mathsf{\displaystyle\int\! \frac{1}{(sin\,x)^2}\cdot cos\,x\,dx\qquad\quad(i)}$


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:  indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus