Integrate using the table of integrals 1 int frac1x2 25 dx 2...

Integrate using the table of integrals. 1. $$\int \frac{1}{x^2-25} dx$$ 2. $$\int \frac{4x^3}{\sqrt{x^8+1}} dx$$ Hint: Let u = x⁴ so u² = x⁸. Evaluate each improper integral or demonstrate that it is divergent. 3. $$\int_0^\infty e^{-3x} dx$$ 4. $$\int_4^\infty x^{-1/2} dx$$

Transcript

00:01 For this question, we need to find the definite integral from pi over 4 to pi over 3 of dx divided by cosine squared of x tangent of x.

00:11 So in order to solve this, i'm going to start by rewriting the integrand using trig identities.

00:16 In order to do this, recall the trig identity that tells us that secant of x is the same thing as 1 over cosine of x.

00:26 This allows us to rewrite integral as the integral from pi over 4 to pi over 3 of secant squared of x divided by tangent of x, bx.

00:42 We're going to use the substitution method by introducing a new variable u to take place of one of the expressions in the integrand.

00:51 So in this case i'm going to let u equal tangent of x that means that the derivative of u du is equal to secant squared of x dx so now for the substitution i can plug in du in place of secant square root of x dx and i can plug in u in place of tangent of x.

01:18 So this gives us a new integrand of 1 over u du and now since i've changed the variable to u i do also need to change the limits of integration.

01:31 So i know that u is equal to tangent of x because that's what i defined u to be.

01:38 So in order to find the new limits of integration, plug in pi over 4 and pi over 3 in place of x.

01:46 So i have tangent of pi over 4 and also tangent of pi over 3.

01:53 So that gives us new limits of integration of 1 and square root of 3.

02:00 Go ahead and write those down as the new limits of integration.

02:03 Integration and now i can find the antiderivative of 1 over u du...