Integral Master's Please Help Me?

Question

How do you solve this problems using the chain rule:

1. the integral of 1/ (1+(5x)^2) dx
2. the integral of square root of (2/x) dx

Answer 1

To solve for #1, you need to use trigonometric substitution. When doing trig substitution, you have to note the form of it. They can be in one of these forms:
a) 1 - [junk]^2
b) [junk]^2 -1
c) 1 + [junk]^2 or [junk]^2 + 1

The above cases mirror the following trig identities.
a) 1 - (cos(x))^2 = (sin(x))^2
b) (sec(x))^2 - 1 = (tan(x))^2
c) 1 + (tan(x))^2 = (sec(x))^2

When doing trig substitution, you have to keep this in mind.

The appropriate trig substitution in #1 is c). You'll want to let x=1/(sqrt(5)tany) and work from there. The answer is much too huge to put on here, and my goal was to get you started anyway.

To solve #2, Note that sqrt(2/x) is the same as sqrt(2)/sqrt(x).

Integral (sqrt(2/x))dx = Integral(sqrt(2)/sqrt(x))dx
= sqrt(2) Integral (1/sqrt(x))dx (I just pulled out the constant).

And then from here, all you have to do is note that 1/sqrt(x) is the same as 1/(x^(1/2)) which is the same as x^(-1/2). Now it's the matter of using the reverse power rule.

sqrt(2) * Integral (x^(-1/2)) = sqrt(2) * 2 * x^(1/2) + C

Answer 2

so (1) y=1/(1+(5x)^2), assume tg(t)=5x, then dx=(1/5)*dt/(cos(t))^2 and y=(cos(t))^2
thus y*dx=0.2*dt, and I(t)=0.2*t+C=0.2*arctan(5x)+C
so (2) y*dx = sqrt(2/x)*dx = sqrt(2)*x^(-1/2)*dx; Mind! (x^n)’=n*x^(n-1).
Thus I= sqrt(2)*x^(-1/2+1)/(-1/2+1) = sqrt(2x)*2