How to find integral of $sinx/cos^2x$ ?

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Answer 1 · 2017-09-29T06:09:03

$intsinx/(cos^2x)dx=secx$

To find $intsinx/(cos^2x)dx$ , let $t=cosx$ , then as $dt=-sinxdx$

$intsinx/(cos^2x)dx=int(-dt)/t^2$

= $-(t^(-1))/(-1)$

= $1/t$

= $1/cosx$

= $secx$

Observe that $d/(dx)secx=secxtanx=sinx/cos^2x$

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