Solved on Feb 08, 2024

Evaluate the indefinite integral $\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x =$ C$

STEP 1

Assumptions
1. We are given the indefinite integral $\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x$ to evaluate.
2. We will use substitution to simplify the integral.
3. We will use the fact that the derivative of $\arccos(u)$ with respect to $u$ is $\frac{-1}{\sqrt{1-u^2}}$ .
4. We will use the constant $C$ to represent the constant of integration.

STEP 2

Let's use substitution to simplify the integral. We will let $u = \arccos(7x)$ . This means that $x = \frac{\cos(u)}{7}$ .

STEP 3

We need to find $dx$ in terms of $du$ . To do this, we differentiate both sides of $x = \frac{\cos(u)}{7}$ with respect to $u$ .
$\frac{dx}{du} = -\frac{1}{7}\sin(u)$

STEP 4

Now we solve for $dx$ .
$dx = -\frac{1}{7}\sin(u) du$

STEP 5

We substitute $x = \frac{\cos(u)}{7}$ and $dx = -\frac{1}{7}\sin(u) du$ into the integral.
$\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x = \int \frac{2 u}{\sqrt{1-49 \left(\frac{\cos(u)}{7}\right)^2}} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 6

Simplify the expression inside the square root.
$1-49 \left(\frac{\cos(u)}{7}\right)^2 = 1 - \cos^2(u)$

STEP 7

Recognize that $1 - \cos^2(u)$ is equal to $\sin^2(u)$ .
$1 - \cos^2(u) = \sin^2(u)$

STEP 8

Substitute $\sin^2(u)$ back into the integral.
$\int \frac{2 u}{\sqrt{\sin^2(u)}} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 9

Simplify the square root of $\sin^2(u)$ to $\sin(u)$ .
$\int \frac{2 u}{\sin(u)} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 10

The $\sin(u)$ terms cancel out.
$\int 2 u \left(-\frac{1}{7}\right) du$

STEP 11

Simplify the integral.
$-\frac{2}{7} \int u \, du$

STEP 12

Integrate $u$ with respect to $u$ .
$-\frac{2}{7} \left(\frac{u^2}{2}\right) + C$

STEP 13

Simplify the constant factor.
$-\frac{1}{7} u^2 + C$

STEP 14

Recall that $u = \arccos(7x)$ and substitute it back into the expression.
$-\frac{1}{7} (\arccos(7x))^2 + C$

STEP 15

Write the final answer.
$\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x = -\frac{1}{7} (\arccos(7x))^2 + C$

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Evaluate the indefinite integral $\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x =$ C$

STEP 1

STEP 2

Let's use substitution to simplify the integral. We will let $u = \arccos(7x)$ . This means that $x = \frac{\cos(u)}{7}$ .

STEP 3

We need to find $dx$ in terms of $du$ . To do this, we differentiate both sides of $x = \frac{\cos(u)}{7}$ with respect to $u$ .
$\frac{dx}{du} = -\frac{1}{7}\sin(u)$

STEP 4

Now we solve for $dx$ .
$dx = -\frac{1}{7}\sin(u) du$

STEP 5

We substitute $x = \frac{\cos(u)}{7}$ and $dx = -\frac{1}{7}\sin(u) du$ into the integral.
$\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x = \int \frac{2 u}{\sqrt{1-49 \left(\frac{\cos(u)}{7}\right)^2}} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 6

Simplify the expression inside the square root.
$1-49 \left(\frac{\cos(u)}{7}\right)^2 = 1 - \cos^2(u)$

STEP 7

Recognize that $1 - \cos^2(u)$ is equal to $\sin^2(u)$ .
$1 - \cos^2(u) = \sin^2(u)$

STEP 8

Substitute $\sin^2(u)$ back into the integral.
$\int \frac{2 u}{\sqrt{\sin^2(u)}} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 9

Simplify the square root of $\sin^2(u)$ to $\sin(u)$ .
$\int \frac{2 u}{\sin(u)} \left(-\frac{1}{7}\sin(u)\right) du$

STEP 10

The $\sin(u)$ terms cancel out.
$\int 2 u \left(-\frac{1}{7}\right) du$

STEP 11

Simplify the integral.
$-\frac{2}{7} \int u \, du$

STEP 12

Integrate $u$ with respect to $u$ .
$-\frac{2}{7} \left(\frac{u^2}{2}\right) + C$

STEP 13

Simplify the constant factor.
$-\frac{1}{7} u^2 + C$

STEP 14

Recall that $u = \arccos(7x)$ and substitute it back into the expression.
$-\frac{1}{7} (\arccos(7x))^2 + C$

STEP 15

Write the final answer.
$\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x = -\frac{1}{7} (\arccos(7x))^2 + C$

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Evaluate the indefinite integral ∫2arccos⁡(7x)1−49x2dx=\int \frac{2 \arccos (7 x)}{\sqrt{1-49 x^{2}}} d x = ∫1−49x2​2arccos(7x)​dx=C$

STEP 1

STEP 2

Let's use substitution to simplify the integral. We will let u=arccos⁡(7x)u = \arccos(7x)u=arccos(7x). This means that x=cos⁡(u)7x = \frac{\cos(u)}{7}x=7cos(u)​.

STEP 3

We need to find dxdxdx in terms of dududu. To do this, we differentiate both sides of x=cos⁡(u)7x = \frac{\cos(u)}{7}x=7cos(u)​ with respect to uuu.dxdu=−17sin⁡(u)\frac{dx}{du} = -\frac{1}{7}\sin(u)dudx​=−71​sin(u)

STEP 4

Now we solve for dxdxdx.dx=−17sin⁡(u)dudx = -\frac{1}{7}\sin(u) dudx=−71​sin(u)du

STEP 5

STEP 6

Simplify the expression inside the square root.1−49(cos⁡(u)7)2=1−cos⁡2(u)1-49 \left(\frac{\cos(u)}{7}\right)^2 = 1 - \cos^2(u)1−49(7cos(u)​)2=1−cos2(u)

STEP 7

Recognize that 1−cos⁡2(u)1 - \cos^2(u)1−cos2(u) is equal to sin⁡2(u)\sin^2(u)sin2(u).1−cos⁡2(u)=sin⁡2(u)1 - \cos^2(u) = \sin^2(u)1−cos2(u)=sin2(u)

STEP 8

Substitute sin⁡2(u)\sin^2(u)sin2(u) back into the integral.∫2usin⁡2(u)(−17sin⁡(u))du\int \frac{2 u}{\sqrt{\sin^2(u)}} \left(-\frac{1}{7}\sin(u)\right) du∫sin2(u)​2u​(−71​sin(u))du

STEP 9

Simplify the square root of sin⁡2(u)\sin^2(u)sin2(u) to sin⁡(u)\sin(u)sin(u).∫2usin⁡(u)(−17sin⁡(u))du\int \frac{2 u}{\sin(u)} \left(-\frac{1}{7}\sin(u)\right) du∫sin(u)2u​(−71​sin(u))du

STEP 10

The sin⁡(u)\sin(u)sin(u) terms cancel out.∫2u(−17)du\int 2 u \left(-\frac{1}{7}\right) du∫2u(−71​)du

STEP 11

Simplify the integral.−27∫u du-\frac{2}{7} \int u \, du−72​∫udu

STEP 12

Integrate uuu with respect to uuu.−27(u22)+C-\frac{2}{7} \left(\frac{u^2}{2}\right) + C−72​(2u2​)+C

STEP 13

Simplify the constant factor.−17u2+C-\frac{1}{7} u^2 + C−71​u2+C

STEP 14

Recall that u=arccos⁡(7x)u = \arccos(7x)u=arccos(7x) and substitute it back into the expression.−17(arccos⁡(7x))2+C-\frac{1}{7} (\arccos(7x))^2 + C−71​(arccos(7x))2+C

STEP 15