integrate 0 to pi/2 xsinx/1+ cos2x

Question

definite integral question

Answer 1

I'm going to rewrite the function with parentheses to what I assume it's supposed to look like:

 

∫(x*sin(x))/(1+cos(2x)) dx, 0, pi/2)

 

It's helpful to know trig identities.

 

cos(2x) = cos(x+x) = cos^2(x) - sin^2(x)

 

Take a look at just the denominator:

 

1 + cos^2(x) - sin^2(x) = (1 - sin^2(x)) + cos^2(x) = cos^2(x) + cos^2(x) = 2*cos^2(x)

 

So now you have:

 

∫x*sin(x))/(2*cos^2(x)) dx = 1/2∫x*sin(x))/(cos^2(x)) dx

 

Let's rewrite this as:

 

1/2∫(sin(x)/(cos(x))*(x/cos(x)) dx

 

1/cos(x) is sec(x)

sin(x)/cos(x) = tan(x)

 

So now we've got:

 

1/2∫x*sec(x)tan(x) dx

 

This would be easy to evaluate if that pesky little x term wasn't there... fortunately, though, we can integrate by parts.

 

Let's set

 

u = x, and dv = sec(x)tan(x)

 

So,

 

du = 1*dx, v = sec(x)

 

And plug in our values for:

 

u*v - ∫v*du

 

We arrive at:

 

x*sec(x) - ∫sec(x) dx

 

Which yields:

 

x*sec(x) - ln(tan(x) + sec(x))

 

Don't forget the 1/2 term:

 

1/2*[x*sec(x) - ln(tan(x) + sec(x))]

 

You'll have to check your upper limit because pi/2, when plugged into both sec(x) and tan(x), results in 1/0 which is undefined. You'll also be able to check that from your original equation, since (1 + cos(2*pi/2)) = (1 + cos(pi)) = 1 - 1 = 0. Since that 0 is in the denominator, the function is undefined.

 

This integral is improper since it approaches infinity at its upper limit.