Question #5a465

1 Answer
NickTheTurtle
Jan 12, 2018

#int\ sin(x)/cos^2(x)\ dx=1/cos(x)+C=sec(x)+C#

Explanation:

#int\ sin(x)/cos^2(x)\ dx#

Let's substitute #t=cos(x)# and #dt/dx=-sin(x)#.

Then, we have
#int\ sin(x)/cos^2(x)\ dx#
#=int\ sin(x)/cos^2(x)dx/dt\ dt#
#=-int\ sin(x)/t^2 1/sin(x)\ dt#
#=-int\ dt/t^2#
#=1/t+C#

Substitute back #t=cos(x)# to get the final answer of
#=1/cos(x)+C#

Another way to solve the integration problem is to use the fact that #tan(x)=sin(x)/cos(x)# and #sec(x)=1/cos(x)# to get that
#int\ sin(x)/cos^2(x)\ dx#
#=int\ tan(x)sec(x)\ dx#

Recall that #d/dx(sec(x))=tan(x)sec(x)#. So, we have the above integral equal to
#=sec(x)+C#

As seen, both two methods get the same answer.

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