Use the comparison method to show that the integral diverges.?

To show that this integral diverges, note that sin²x ≥ 0 for all x.

So, (1 + sin²x)/√x ≥ (1 + 0)/√x = x^(-1/2) for all x > 0.

Since ∫(x = 1 to ∞) x^(-1/2) dx = 2x^(1/2) {for x = 1 to ∞} = ∞ (divergent), we conclude that the original integral diverges by the Comparison Test.

I hope this helps!

int_1^infinity (1 + sin^2(x))/sqrt(x) dx

Let y = 2sqrt(x) dx; dy = 1/sqrt(x) dx

int_2^(infinity) [1 + sin^2((y^2)/4)] dy = diverges

From 1 to 8 the function is continuous and well defined. So the integral from 1 to 8 is finite; it converges. it does not diverge.

To your update, hint: Compare it to 1/√x

hard to do ...since it converges......( 1 + sin² x ) / √ x ≤ 2 / √ x...and int over x in [ 1 , 8 ] of {2 / √x dx } is 8√2 - 4....if you meant over [ 1 , ∞ ) then it does diverge since 1 / √x ≤ ( 1 + sin² x ) / √x