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Evaluate the indefinite integral:
[tex]\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)}[/tex]
Make the following substitution:
[tex]\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}[/tex]
and then, the integral (i) becomes
[tex]=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
=\mathsf{\displaystyle\int\! u^{-2}\,du}[/tex]
Integrate it by applying the power rule:
[tex]\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
\mathsf{=\dfrac{u^{-1}}{-1}+C}\\\\\\
\mathsf{=-\,\dfrac{1}{u}+C}[/tex]
Now, substitute back for u = sin x, so the result is given in terms of x:
[tex]\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}[/tex]
[tex]\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark[/tex]
I hope this helps. =)
Tags: indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus
