Math  /  Calculus

QuestionFor questions 3-5, determine the indefinite integral. Check your work by differentiation
3. (sec2x8)dx\int\left(\sec ^{2} x-8\right) d x \% x2cosdx8dx\int x^{2} \cos d x-\int 8 d x 8tanx8x8 \tan x-8 x
4. 110ydy\int \frac{1}{10 y} d y 1100dy\int \frac{1}{100} d y $10lny\$ 10 \ln |y|
5. (2s+7)2ds\int(2 s+7)^{2} d s (4s2+28s+4(9)dsxn1424s2ds+28sds+449ds\begin{array}{l} \int\left(4 s^{2}+28 s+4(9) d s\right. \\ \int x^{n^{-1}} \\ \int \frac{4}{2}-4 s^{2} d s+28 \int s d s+4 \int 49 d s \end{array}

Studdy Solution

STEP 1

What is this asking? We need to find the indefinite integrals of three functions and then double-check our answers by taking the derivative of each result. Watch out! Don't forget to add the constant of integration, +C+C, because the derivative of a constant is always zero!
Also, remember the power rule for integration carefully.

STEP 2

1. Integrate secant squared minus 8
2. Integrate one over ten y
3. Integrate two s plus 7 squared

STEP 3

Let's **break down** this integral!
We have (sec2(x)8)dx\int (\sec^2(x) - 8) \, dx.
We can split this into two **easier integrals**: sec2(x)dx\int \sec^2(x) \, dx and 8dx\int -8 \, dx.

STEP 4

Now, what's the integral of sec2(x)\sec^2(x)?
It's tan(x)\tan(x)!
Why? Because the **derivative** of tan(x)\tan(x) is sec2(x)\sec^2(x).
It all comes back around!

STEP 5

Next up, 8dx\int -8 \, dx.
The integral of a **constant** is just that constant times xx, so we get 8x-8x.

STEP 6

**Putting it all together**, we have tan(x)8x\tan(x) - 8x.
And don't forget our **constant of integration**, +C+C!
So, our **final answer** is tan(x)8x+C\tan(x) - 8x + C.

STEP 7

Let's **check our work**!
The derivative of tan(x)\tan(x) is sec2(x)\sec^2(x), the derivative of 8x-8x is 8-8, and the derivative of a constant is 00.
Adding those up, we get sec2(x)8\sec^2(x) - 8, which is exactly what we started with!
Awesome!

STEP 8

We have 110ydy\int \frac{1}{10y} \, dy.
We can **rewrite** this as 1101ydy\frac{1}{10} \int \frac{1}{y} \, dy.

STEP 9

Now, the integral of 1y\frac{1}{y} with respect to yy is lny\ln|y|.
The absolute value is **super important** because we can't take the logarithm of a negative number!

STEP 10

So, our integral becomes 110lny\frac{1}{10} \ln|y|.
Don't forget the **constant of integration**, giving us a **final answer** of 110lny+C\frac{1}{10} \ln|y| + C.

STEP 11

Time to **check**!
The derivative of 110lny\frac{1}{10} \ln|y| is 1101y=110y\frac{1}{10} \cdot \frac{1}{y} = \frac{1}{10y}, and the derivative of CC is 00.
We got back to the **original function**, so we're good!

STEP 12

We're looking at (2s+7)2ds\int (2s + 7)^2 \, ds.
First, let's **expand** that square: (2s+7)2=(2s+7)(2s+7)=4s2+14s+14s+49=4s2+28s+49(2s + 7)^2 = (2s + 7)(2s + 7) = 4s^2 + 14s + 14s + 49 = 4s^2 + 28s + 49.

STEP 13

So, our integral becomes (4s2+28s+49)ds\int (4s^2 + 28s + 49) \, ds.
We can **split this up** into 4s2ds+28sds+49ds\int 4s^2 \, ds + \int 28s \, ds + \int 49 \, ds.

STEP 14

Using the **power rule** for integration, 4s2ds=4s33\int 4s^2 \, ds = \frac{4s^3}{3}, 28sds=28s22=14s2\int 28s \, ds = \frac{28s^2}{2} = 14s^2, and 49ds=49s\int 49 \, ds = 49s.

STEP 15

**Combining** these, we get 4s33+14s2+49s\frac{4s^3}{3} + 14s^2 + 49s.
And, of course, we add our **constant of integration**, +C+C, for a **final answer** of 4s33+14s2+49s+C\frac{4s^3}{3} + 14s^2 + 49s + C.

STEP 16

Let's **double-check**!
Taking the derivative, we get 4s2+28s+49+04s^2 + 28s + 49 + 0, which simplifies to (2s+7)2(2s + 7)^2.
Perfect!

STEP 17

1. tan(x)8x+C\tan(x) - 8x + C
2. 110lny+C\frac{1}{10} \ln|y| + C
3. 4s33+14s2+49s+C\frac{4s^3}{3} + 14s^2 + 49s + C

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