The integral of 3xsec^2(x) dx

Question

I'm not sure how to start this one. All I know is sec^2 is (1/cosx)^2 is this right?

Answer 1

We can rewrite the integral so that the constant is outside the integral.

 

3∫xsec²xdx

 

 

For this one, we use the integration by parts method.  We let u be a factor we can derivative, and dv be a factor we easily integrate.

 

Let the following:

 

u = x                                                              dv = sec²xdx

du = dx                                                            v = tanx

 

3∫xsec²xdx = 3[xtanx - ∫tanxdx]

 

 

Then knowing that    tanx = sinx/cosx

 

3∫xsec²xdx = 3[xtanx - ∫sinx(cosx)^-1dx]

 

 

Use the substitution method for the subintegral   ∫sinx(cosx)^-1dx.

 

   z = cosx

 dz = -sinxdx

-dz = sinxdx

 

Using this substitution, the subintegral part is

 

-∫z^-1dz = -ln(z) = -ln(cosx)

 

 

Now plugging in this subintegral back into the original integral, we get

 

3∫xsec²xdx = 3[xtanx - (-ln(cosx))] + C

 

3∫xsec²xdx = 3[xtanx + ln(cosx)] + C

 

 

Now we check this solution by taking its derivative.  We should get back the integral.

 

d/dx[3(xtanx + ln(cosx)] =

 

3((tanx + xsec²x) + (-sinx / cosx)) =

 

3(tanx + xsec²x - tanx) =

 

3xsec²x

 

 

This checks out.