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Integrate the function $\frac{\sin\left(x\right)}{1+\cos\left(x\right)}$ from $\frac{3\pi }{2}$ to $2\pi $

Step-by-step Solution

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Final answer to the problem

$-\ln\left(2\right)$
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Step-by-step Solution

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1

Simplifying

$\int_{\frac{3\pi}{2}}^{2\pi }\frac{\sin\left(x\right)}{1+\cos\left(x\right)}dx$
2

We can solve the integral $\int\frac{\sin\left(x\right)}{1+\cos\left(x\right)}dx$ by applying the method Weierstrass substitution (also known as tangent half-angle substitution) which converts an integral of trigonometric functions into a rational function of $t$ by setting the substitution

$t=\tan\left(\frac{x}{2}\right)$
3

Hence

$\sin x=\frac{2t}{1+t^{2}},\:\cos x=\frac{1-t^{2}}{1+t^{2}},\:\mathrm{and}\:\:dx=\frac{2}{1+t^{2}}dt$
4

Substituting in the original integral we get

$\int_{\frac{3\pi}{2}}^{2\pi }\frac{\frac{2t}{1+t^{2}}}{1+\frac{1-t^{2}}{1+t^{2}}}\frac{2}{1+t^{2}}dt$
5

Simplifying

$\int_{\frac{3\pi}{2}}^{2\pi }\frac{2t}{1+t^{2}}dt$
6

We can solve the integral $\int_{\frac{3\pi}{2}}^{2\pi }\frac{2t}{1+t^{2}}dt$ by applying integration by substitution method (also called U-Substitution). First, we must identify a section within the integral with a new variable (let's call it $u$), which when substituted makes the integral easier. We see that $1+t^{2}$ it's a good candidate for substitution. Let's define a variable $u$ and assign it to the choosen part

$u=1+t^{2}$
7

Now, in order to rewrite $dt$ in terms of $du$, we need to find the derivative of $u$. We need to calculate $du$, we can do that by deriving the equation above

$du=2tdt$
8

Isolate $dt$ in the previous equation

$\frac{du}{2t}=dt$
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Substituting $u$ and $dt$ in the integral and simplify

$\int_{\frac{3\pi}{2}}^{2\pi }\frac{1}{u}du$
10

The integral of the inverse of the lineal function is given by the following formula, $\displaystyle\int\frac{1}{x}dx=\ln(x)$

$\left[\ln\left(u\right)\right]_{\frac{3\pi}{2}}^{2\pi }$
11

Replace $u$ with the value that we assigned to it in the beginning: $1+t^{2}$

$\left[\ln\left(1+t^{2}\right)\right]_{\frac{3\pi}{2}}^{2\pi }$
12

Replace $t$ with the value that we assigned to it in the beginning: $\tan\left(\frac{x}{2}\right)$

$\left[\ln\left(1+\tan\left(\frac{x}{2}\right)^{2}\right)\right]_{\frac{3\pi}{2}}^{2\pi }$
13

Evaluate the definite integral

$\ln\left(1+\tan\left(\frac{2\pi }{2}\right)^{2}\right)-\ln\left(1+\tan\left(\frac{\frac{3\pi}{2}}{2}\right)^{2}\right)$
14

Simplify the expression inside the integral

$-\ln\left(2\right)$

Final answer to the problem

$-\ln\left(2\right)$

Exact Numeric Answer

$-0.693147$

Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

Solve integral of (sinx/(1+cosx))dx from 3\pi /2 to 2\pi using basic integralsSolve integral of (sinx/(1+cosx))dx from 3\pi /2 to 2\pi using u-substitutionSolve integral of (sinx/(1+cosx))dx from 3\pi /2 to 2\pi using integration by parts

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Function Plot

Plotting: $\frac{\sin\left(x\right)}{1+\cos\left(x\right)}$

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a
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u
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x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Definite Integrals

Given a function f(x) and the interval [a,b], the definite integral is equal to the area that is bounded by the graph of f(x), the x-axis and the vertical lines x=a and x=b

Used Formulas

1. See formulas

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