Determine: (a) ∫2 sin(3 x+1) d x (b) ∫√(5-2 x) d x (c) ∫6 ·e^t-1 x d x (d) ∫(4 x+1)^3 d x (e) ∫

step 1 For this problem, you're asked to solve a ton of integrals. And in looking at all of the integra

Determine: (a) ∫2 sin(3 x+1) d x (b) ∫√(5-2 x) d x (c) ∫6...

Key Concepts

Integration of Power Functions

This involves integrating expressions where a function is raised to a power. When the base of the power is a linear expression or a more general function, substitution is often employed to transform the integral into a basic power rule form. This rule states that the integral of u^n is u^(n+1) divided by (n+1), for n not equal to -1.

Integration Leading to Logarithmic Functions

When dealing with integrals of functions that involve a variable in the denominator in the form 1/(ax+b), the result is often a logarithmic function. Recognizing this structure and using the substitution method can simplify the integration process, yielding an antiderivative that includes the natural logarithm.

Integration of Radical Functions

Integrals involving radical expressions, such as square roots, are managed by rewriting the radical as an exponent. Once in this form, the power rule for integration can be applied. Substitution may be used when the expression under the radical is a function of the variable, thereby reducing the integral to a standard power integral.

Integration of Exponential Functions

Integrating exponential functions requires familiarity with the properties of the exponential function. When the exponent is a linear function or has other modifications, substitution is typically used to simplify the integral into the standard form that integrates to the exponential function divided by the derivative of the inner function.

Integration of Trigonometric Functions

This concept involves knowing the antiderivatives of standard trigonometric functions, such as sine, cosine, and secant squared, which are used to solve integrals involving trigonometric expressions. Mastery of these antiderivatives, as well as the ability to apply substitution when these functions appear with arguments that are linear or otherwise modified, is essential.

u?Substitution

This is a method of integration used when the integrand is a composite function. By substituting a part of the integrand with a new variable, the integral can be rewritten in a simpler form that is easier to evaluate. This technique is essentially the reverse application of the chain rule from differentiation.

Transcript

00:01 For this problem, you're asked to solve a ton of integrals.

00:07 And in looking at all of the integrals, i see that these are all in the form of f of g of x.

00:14 And so we'll need to use the substitution rule in order to solve, which essentially says that if you have the integral of f of g of x, g prime of x, d .x, you can substitute u for g of x and solve the integral f of x.

00:30 Sorry, f of u, d, u, and then substitute back in u at the end in order to solve.

00:37 So what does that look like in practice? i think it's really, really helpful to actually identify what u and d .u are going to be.

00:48 So u in this case is going to be what's inside of the sign.

00:53 So we have 3x plus 1.

00:57 Excuse me, makes du equal to 3.

01:02 Dx and to make our lives a little bit easier i'm going to solve for dx here.

01:13 Solve in quotation marks because that is not really what's happening but mathematically it works.

01:20 So we end up here with one third d .u is equal to dx and now we can go ahead and substitute that into our original integral.

01:33 So we would end up with two times the integral of sine of u times one -third d u.

01:46 So i'm going to bring the one -third out.

01:49 We end up with two -thirds, the integral of sine u -d -u.

01:57 And the integral of sign is one that we can easily know.

02:02 It's minus cosine.

02:03 So now we end up with negative two -thirds, of u plus c and then we have to of course plug you back in so our final answer is negative two -thirds cosine of three x plus one plus c so that is the is sort of the idea we'll go ahead and try with us the rest of the problems here but again i think the important thing is identifying what u and d, you are going to be, and then you can go ahead and solve integration normally from there once you've done your substitution.

02:49 So b is the integral of the square root of 5 minus 2x, dx.

03:02 I identified u to be 5 minus 2x, which means that du is minus 2dx.

03:16 So when i substitute, i end up with minus 1 half, the integral of, i'm going to write it as u to the minus 1 half, du.

03:31 And we can go ahead and integrate that using the power rule, and you would end up with minus 1⁄2.

03:40 Again, let's plug our u value back in.

03:46 So 5 minus 2x to the three halves all over three halves plus c.

03:56 And we can go ahead and further simplify that to minus one -third times five minus two x to the three - halves plus c.

04:11 We've done two examples now.

04:15 Now and there are quite a few more to go but they're all going to be the same idea of using the substitution rule so let's look at c for c we have the integral of 6 e to the t minus x dx i think there was a typo because this is kind of strange but we're going to go ahead and run with it and i set u equal to t minus x, which makes d u equal to negative dx.

04:59 So we are integrating here, negative 6 integral of e to the u, du, which ends up just being negative 6, e to the u, or e to the t minus x, plus.

05:18 Plus c...

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