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Evaluate the indefinite integral. (Use C for the constant of integration.) sin(10x) 1 cos2(10x) dx

Schach [20]

Answer:

The result of the integral is $-\frac{\cos^3{10x}}{30} + C$

Step-by-step explanation:

We are given the following integral:

$\int \sin{(10x)}\cos^{2}{(10x)} dx$

I am going to solve by substitution, using $u = \cos{10x}$ , so $du = -10\sin{10x}dx, dx = -\frac{du}{10\sin{10x}}$ . So, we have

$-\frac{1}{10} \int u^2 du$

Which has the following result:

$-\frac{u^3}{30} + C$

Going back to x, the result of the integral is

$-\frac{\cos^3{10x}}{30} + C$

5 0

3 years ago

Y=5+3x for x <br> Show work

Levart [38]

Answer:

$x = \frac{y - 5}{3}$

Step-by-step explanation:

Hello!

To solve for x, we have to first isolate the variable.

<h3>Solve for x</h3>

$y = 5 + 3x$
$y - 5 = 3x$
$y - 5 = 3(x)$
$\frac{y - 5}{3} = x$

The value of x is $\frac{y - 5}{3}$ .

5 0

2 years ago

Find the exact values of sin θ/2 and cos θ/2 for sin θ = 3/4 on the interval 0° ≤ θ ≤ 90°

timofeeve [1]

Answer:

Step-by-step explanation:

If our interval is between 0 and 90 degrees, that means that our angles are first quadrant angles. Since the formulas for the half angles of sin and cos consider both the positive and negative roots, knowing that we are in QI leaves us using only the principle, or positive, root.

Also, if we know that the sin of angle is 3/4, we can set up a right triangle in QI and find the missing side of that triangle using Pythagorean's Theorem. Sin of an angle is the ratio opposite side/hypotenuse. That means that, according to Pythagorean's Theorem,

$4^2=3^2+b^2$ and

$16-9=b^2$ and

$b=\sqrt{7}$

Now for the forula for the sin of a half-angle:

$sin(\frac{\theta}{2})=\sqrt{\frac{1-cos\theta}{2} }$

Referring to the right triangle we created and found the missing side for, we see that the cosine of that same triangle is $\frac{\sqrt{7} }{4}$

Filling in the formula then:

$sin(\frac{\theta}{2})=\sqrt{\frac{1-\frac{\sqrt{7} }{4} }{2} }$

Get a common denominator on the upper fraction:

$sin(\frac{\theta}{2})=\sqrt{\frac{\frac{4-\sqrt{7} }{4} }{2} }$

Bring up the 2 as 1/2 to multiply and get:

$sin(\frac{\theta}{2})=\sqrt{\frac{4-\sqrt{7} }{8} }$

When you do the exact same thing to find the cos of the half angle, you are using a + sign instead of a - sign under the radical. That gives you as your exact value:

$cos(\frac{\theta}{2})=\sqrt{\frac{4+\sqrt{7} }{8} }$