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Evaluate the following integrals. a) \int \frac{\sin 20t}{\sin 10t} dt b) \int _{0}^{3} (6+...

Question:

Evaluate the following integrals.

a) {eq}\int \frac{\sin 20t}{\sin 10t} dt {/eq}

b) {eq}\int _{0}^{3} (6+ 6y -y^2)dy {/eq}

c) {eq}\int (\frac{1-x}{x})^2 dx {/eq}

Integration:

Let us consider a function f(t) where t is time, then the integral to that function is found out as follows:

{eq}\displaystyle \int f(t) \, dt = F(t) + C {/eq}.

From the above equation, {eq}F(t) {/eq} is the integral and C is an arbitrary constant.

Sum Rule:

{eq}\displaystyle \int f(x) + g(x) dx = \int f(x) dx + \int g(x) dx {/eq}

Formulas used are:

{eq}\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} {/eq}

{eq}\displaystyle \int \sin x dx = -\cos x {/eq}

{eq}\displaystyle \int \cos x dx = \sin x {/eq}

let us consider a function {eq}f(x) {/eq} then

{eq}\displaystyle \int \frac{f'(x)}{f(x)}dx = \ln |f(x)| {/eq}

{eq}\displaystyle \sin (2x) = 2\sin x\cos x {/eq}

Answer and Explanation:


a.

Given function is

{eq}\displaystyle \int \frac{\sin (20t)}{\sin (10t)}dt {/eq}

{eq}\displaystyle =\int \frac{\sin (2(10t))}{\sin (10t)}dt {/eq}

By using {eq}\sin (2x)=2\sin x\cos x {/eq} in above equation we get

{eq}\displaystyle =\int \frac{2\sin (10t)\cos (10t)}{\sin (10t)}dt {/eq}

{eq}\displaystyle =2\int \cos (10t)dt {/eq}

{eq}\displaystyle =\frac{2\sin (10t)}{10} {/eq}

The answer is

{eq}\displaystyle \int \frac{\sin (20t)}{\sin (10t)}dt = \frac{\sin (10t)}{5}+c {/eq}

where c is any constant.


b.

Given function is

{eq}\displaystyle \int_0^3 (6+6y-y^2)dy {/eq}

By using sum rule above function can be written as

{eq}\displaystyle =\int_0^3 6dy + \int_0^3 6ydy - \int_0^3 y^2dy {/eq}

{eq}\displaystyle =\left[6y\right]_0^3 + \left[3y^2\right]_0^3 - \left[\frac{y^3}{3}\right]_0^3 {/eq}

{eq}\displaystyle =(18-0) + (27-0) - (9-0) {/eq}

{eq}\displaystyle =36 {/eq}

The answer is

{eq}\displaystyle \int_0^3 (6+6y-y^2)dy = 36 {/eq}


c.

Given function is

{eq}\displaystyle \int (\frac{1-x}{x})^2 dx {/eq}

{eq}\displaystyle \int (\frac{1-2x+x^2}{x^2})dx {/eq}

{eq}\displaystyle =\int (\frac{1}{x^2} -2\frac{1}{x} +1)dx {/eq}

By using sum rule above equation can be written as

{eq}\displaystyle =\int \frac{1}{x^2}dx - 2\int \frac{dx}{x} + \int 1dx {/eq}

{eq}\displaystyle =\frac{-1}{x}-2\ln |x| +x {/eq}

The answer is

{eq}\displaystyle \int (\frac{1-x}{x})^2 dx=\frac{-1}{x}-2\ln |x| +x+c {/eq}


Learn more about this topic:

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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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