Find the antiderivative of ..?

#8cosxtanx#

2 Answers
Jan 26, 2018

#int8cosxtanxdx=-8cosx+"C"#

Explanation:

Given: #int8cosxtanxdx#

Take out the constant #8# and rewrite #tanx# as #sinx/cosx# to get the following integral:

#8intcosxsinx/cosxdx#

The cosines cancel and we are left with this simple integral:

#8*intsinxdx#

The integral #intsinxdx# is a common integral whose antiderivative is #-cosx+"C"#

So

#8*intsinxdx=8*-cosx+"C"=-8cosx+"C"#

Jan 26, 2018

The answer is #-8cos(x)+C#.

Explanation:

Remember that the definition of #tan(theta)# is #sin(theta)/cos(theta)#.

#int8cos(x)*tan(x)*dx#

#8*intcancel(cos(x))*sin(x)/cancel(cos(x))*dx#

#8*intsin(x)#

#8*-cos(x)#

#-8cos(x)+C#