(a) Find the derivatives of sin(x^2+1) and sin(x^3+1) (b) Use your answer to part (a) to find an

(a) Find the derivatives of sin(x^2+1) and sin(x^3+1) (b)...

(a) Find the derivatives of $\sin \left(x^{2}+1\right)$ and $\sin \left(x^{3}+1\right)$ (b) Use your answer to part (a) to find antiderivatives of: (i) $x \cos \left(x^{2}+1\right)$ (ii) $x^{2} \cos \left(x^{3}+1\right)$ (c) Find the general antiderivatives of: (i) $x \sin \left(x^{2}+1\right)$ (ii) $x^{2} \sin \left(x^{3}+1\right)$

(a) Find the derivatives of $\sin \left(x^{2}+1\right)$ and $\sin \left(x^{3}+1\right)$
(b) Use your answer to part (a) to find antiderivatives of:
(i) $x \cos \left(x^{2}+1\right)$
(ii) $x^{2} \cos \left(x^{3}+1\right)$
(c) Find the general antiderivatives of:
(i) $x \sin \left(x^{2}+1\right)$
(ii) $x^{2} \sin \left(x^{3}+1\right)$

Key Concepts

Chain Rule

The chain rule is a fundamental differentiation technique used for calculating the derivative of a composite function. In this method, the derivative of the outer function is evaluated at the inner function, and then multiplied by the derivative of the inner function. It is particularly useful when differentiating functions like sine or cosine that have a polynomial or more complex expression inside.

Integration by Substitution

Integration by substitution is the counterpart to the chain rule in integration. It simplifies the process of finding antiderivatives of composite functions by replacing a complicated expression with a single variable. This method involves identifying a part of the integrand as a new variable such that its derivative is also present in the integrand, making the integration process more straightforward.

Transcript

00:01 So today we're going to be doing a problem that involves taking the derivative and anti -derivative of different trig functions.

00:08 So part a is asking us to take the derivative of these two functions.

00:13 And when you take a derivative of a trig function, it's similar to the chain rule, where you take the derivative of the outside, which in this case is cosine, keeping the inside the same, then multiply it by the derivative of the inside, which in this case is 2x.

00:35 You can write the 2x in the front, just to make it a little nicer and match the answer in the textbook.

00:43 Very similar process with this second problem.

00:47 Derivative of sine is cosine.

00:50 Leaving the inside the same, then multiply it by the derivative of x cubed, since that one, that constant, this goes to zero, which is 3x squared.

01:04 Once again, writing it in the front, just to look a little nicer.

01:12 That's for part a.

01:13 Part b, it's asking us to use those answers to find the anti -derivatives of these two given terms.

01:21 Just to make it a little easier, i wrote the answers to our previous part a over here.

01:28 So i was asking us to find the antiderative of x times cosine x squared plus 1.

01:35 But we already found the derivative of sine of x squared plus 1 is 2x times cosine x squared plus 1.

01:44 This is very similar to this, but the only difference is this has a plus or has a 2 in the front.

01:53 So you're going to need to find a term that has, when you take the anti -derivative of it, it's going to give you, instead of 2x, it's just going to give you x.

02:08 So just sign x squared plus one with no coefficients in the front gives you two times what's asking us to find the anti -derivative of.

02:18 So to make that two not present in the answer, we're gonna wanna have a one half the front.

02:27 So imagine if you take the derivative just this like we did in the previous problem, but have this one half in the front.

02:34 So two times this one half will get you this just an x without that two in the front.

02:42 Similar process with this problem.

02:44 The only difference between this and this is this three in the front...

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