How do you integrate #int x/ cos^2(x) dx#?

Question

How do you integrate #int x/ cos^2(x)  dx#?

Christopher Allers · Accepted Answer

I found #xtan(x)+lnabs(cosx)+C#

After using a basic trig identity, this integral becomes a simple case of integration by parts.

Note that #x/cos^2(x)=x*1/cos^2(x)=xsec^2(x)#. That means we can rewrite the integral as #intxsec^2(x)dx#. Because #x# and #sec^2(x)# are being multiplied, we use integration by parts, which tells us: #intudv=uv-intvdu#

In other words, we make a relatively complicated integral like #intxsec^2(x)dx# simpler. The only difficulty is choosing our #u# and #dv#. In this case, #u=x# and #dv# equals everything else, namely #sec^2(x)dx# (when dealing with algebra and trig functions in integration by parts, always choose the algebraic function as your #u#). Now we find #du# and #v#: #u=x=>(du)/dx=1=>du=dx#

#dv=sec^2(x)dx=>intdv=intsec^2(x)dx=>v=tan(x)# We can ignore the constant of integration in this case.

Now we know #u=x#, #du=dx#, #v=tan(x)#. Plugging these in to the integration by parts formula, we have: #intxsec^2(x)dx=xtan(x)-inttan(x)dx#

We could do #inttan(x)dx#, but that would be a waste of time because there is a far easier way: consult a table of integrals! After doing so, we find the result is either #lnabs(sec(x))# or #-lnabs(cos(x))# (both are the same thing). I'm going to use #-lnabs(cos(x))#. #intxsec^2(x)dx=xtan(x)-(-lnabs(cosx))+C# #=xtan(x)+lnabs(cosx)+C#

And that is our final answer.

Aurora Alvarado · Answer

undefined To integrate $ \frac{x}{\cos^2(x)} $ with respect to $ x $, you can use integration by parts method. Let $ u = x $ and $ dv = \frac{1}{\cos^2(x)}dx $.

$$ du = dx $$

$$ v = \int \frac{1}{\cos^2(x)} dx = \tan(x) $$

Now apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

$$ \int \frac{x}{\cos^2(x)} dx = x \tan(x) - \int \tan(x) dx $$

Now integrate $ \tan(x) $ using the formula:

$$ \int \tan(x) dx = -\ln|\cos(x)| + C $$

Thus, the final result is:

$$ \int \frac{x}{\cos^2(x)} dx = x \tan(x) + \ln|\cos(x)| + C $$

where $ C $ is the constant of integration.