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#