Find The Value Of $\int \frac{1}{1+\cosh X} \, Dx$.

Introduction

In mathematics, integration is a fundamental concept that plays a crucial role in solving various problems in calculus, physics, and engineering. One of the most common types of integrals is the trigonometric integral, which involves the integration of functions that contain trigonometric functions such as sine, cosine, and hyperbolic functions. In this article, we will focus on finding the value of the integral $\int \frac{1}{1+\cosh x} \, dx$, which is a classic example of a trigonometric integral.

Understanding the Integral

The given integral is $\int \frac{1}{1+\cosh x} \, dx$. To find the value of this integral, we need to understand the properties of the hyperbolic function $\cosh x$. The hyperbolic cosine function is defined as $\cosh x = \frac{e^x + e^{-x}}{2}$. Using this definition, we can rewrite the integral as $\int \frac{1}{1+\frac{e^x + e^{-x}}{2}} \, dx$.

Simplifying the Integral

To simplify the integral, we can start by combining the terms in the denominator. We have $1 + \frac{e^x + e^{-x}}{2} = \frac{2 + e^x + e^{-x}}{2}$. Therefore, the integral becomes $\int \frac{1}{\frac{2 + e^x + e^{-x}}{2}} \, dx$. To simplify this further, we can multiply the numerator and denominator by 2, which gives us $\int \frac{2}{2 + e^x + e^{-x}} \, dx$.

Using Trigonometric Substitution

To evaluate this integral, we can use the trigonometric substitution $x = 2 \arctan t$. This substitution is useful because it allows us to express the hyperbolic function $\cosh x$ in terms of the trigonometric function $\tan x$. Using this substitution, we have $\cosh x = \frac{e^x + e^{-x}}{2} = \frac{e^{2 \arctan t} + e^{-2 \arctan t}}{2} = \frac{1 + \tan^2 t}{1 + \tan^2 t} = \tan^2 t$.

Evaluating the Integral

Now that we have expressed the hyperbolic function $\cosh x$ in terms of the trigonometric function $\tan x$, we can rewrite the integral as $\int \frac{2}{2 + \tan^2 t} \, dt$. To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\int \frac{2}{2 + u^2} \, du$.

Using Partial Fractions

To evaluate this integral, we can use the method of partial fractions. We can write the integral as $\int \frac{2}{2 + u^2} \, du = \int \frac{A}{u + \sqrt{2}} + \frac{B}{u - \sqrt{2}} \, du$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $2 = A(u - \sqrt{2}) + B(u + \sqrt{2})$. Equating the coefficients of $u$, we get $0 = A + B$. Equating the constant terms, we get $2 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{\sqrt{2}}$ and $B = -\frac{1}{\sqrt{2}}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the integral as $\int \frac{1}{\sqrt{2}} \left( \frac{1}{u + \sqrt{2}} - \frac{1}{u - \sqrt{2}} \right) \, du$. To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1}{\tan t + \sqrt{2}} - \frac{1}{\tan t - \sqrt{2}} \right) \sec^2 t \, dt$.

Simplifying the Integral

To simplify the integral, we can use the trigonometric identity $\sec^2 t = 1 + \tan^2 t$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1 + \tan^2 t}{\tan t + \sqrt{2}} - \frac{1 + \tan^2 t}{\tan t - \sqrt{2}} \right) \, dt$.

Evaluating the Integral

To evaluate this integral, we can use the substitution $v = \tan t$, which gives us $dv = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1 + v^2}{v + \sqrt{2}} - \frac{1 + v^2}{v - \sqrt{2}} \right) \, dv$.

Simplifying the Integral

To simplify the integral, we can use the method of partial fractions. We can write the integral as $\frac{1}{\sqrt{2}} \int \left( \frac{A}{v + \sqrt{2}} + \frac{B}{v - \sqrt{2}} \right) \, dv$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $1 + v^2 = A(v - \sqrt{2}) + B(v + \sqrt{2})$. Equating the coefficients of $v$, we get $0 = A + B$. Equating the constant terms, we get $1 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the integral as $\frac{1}{\sqrt{2}} \int \left( \frac{1/2}{v + \sqrt{2}} - \frac{1/2}{v - \sqrt{2}} \right) \, dv$. To evaluate this integral, we can use the substitution $v = \tan t$, which gives us $dv = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1}{\tan t + \sqrt{2}} - \frac{1}{\tan t - \sqrt{2}} \right) \sec^2 t \, dt$.

Simplifying the Integral

To simplify the integral, we can use the trigonometric identity $\sec^2 t = 1 + \tan^2 t$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1 + \tan^2 t}{\tan t + \sqrt{2}} - \frac{1 + \tan^2 t}{\tan t - \sqrt{2}} \right) \, dt$.

Evaluating the Integral

To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1 + u^2}{u + \sqrt{2}} - \frac{1 + u^2}{u - \sqrt{2}} \right) \, du$.

Simplifying the Integral

To simplify the integral, we can use the method of partial fractions. We can write the integral as $\frac{1}{2\sqrt{2}} \int \left( \frac{A}{u + \sqrt{2}} + \frac{B}{u - \sqrt{2}} \right) \, du$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $1 + u^2 = A(u - \sqrt{2}) + B(u + \sqrt{2})$. Equating the coefficients of $u$, we get $0 = A + B$. Equating the constant terms, we get $1 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the

Introduction

In mathematics, integration is a fundamental concept that plays a crucial role in solving various problems in calculus, physics, and engineering. One of the most common types of integrals is the trigonometric integral, which involves the integration of functions that contain trigonometric functions such as sine, cosine, and hyperbolic functions. In this article, we will focus on finding the value of the integral $\int \frac{1}{1+\cosh x} \, dx$, which is a classic example of a trigonometric integral.

Understanding the Integral

The given integral is $\int \frac{1}{1+\cosh x} \, dx$. To find the value of this integral, we need to understand the properties of the hyperbolic function $\cosh x$. The hyperbolic cosine function is defined as $\cosh x = \frac{e^x + e^{-x}}{2}$. Using this definition, we can rewrite the integral as $\int \frac{1}{1+\frac{e^x + e^{-x}}{2}} \, dx$.

Simplifying the Integral

To simplify the integral, we can start by combining the terms in the denominator. We have $1 + \frac{e^x + e^{-x}}{2} = \frac{2 + e^x + e^{-x}}{2}$. Therefore, the integral becomes $\int \frac{1}{\frac{2 + e^x + e^{-x}}{2}} \, dx$. To simplify this further, we can multiply the numerator and denominator by 2, which gives us $\int \frac{2}{2 + e^x + e^{-x}} \, dx$.

Using Trigonometric Substitution

To evaluate this integral, we can use the trigonometric substitution $x = 2 \arctan t$. This substitution is useful because it allows us to express the hyperbolic function $\cosh x$ in terms of the trigonometric function $\tan x$. Using this substitution, we have $\cosh x = \frac{e^x + e^{-x}}{2} = \frac{e^{2 \arctan t} + e^{-2 \arctan t}}{2} = \frac{1 + \tan^2 t}{1 + \tan^2 t} = \tan^2 t$.

Evaluating the Integral

Now that we have expressed the hyperbolic function $\cosh x$ in terms of the trigonometric function $\tan x$, we can rewrite the integral as $\int \frac{2}{2 + \tan^2 t} \, dt$. To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\int \frac{2}{2 + u^2} \, du$.

Using Partial Fractions

To evaluate this integral, we can use the method of partial fractions. We can write the integral as $\int \frac{2}{2 + u^2} \, du = \int \frac{A}{u + \sqrt{2}} + \frac{B}{u - \sqrt{2}} \, du$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $2 = A(u - \sqrt{2}) + B(u + \sqrt{2})$. Equating the coefficients of $u$, we get $0 = A + B$. Equating the constant terms, we get $2 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{\sqrt{2}}$ and $B = -\frac{1}{\sqrt{2}}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the integral as $\int \frac{1}{\sqrt{2}} \left( \frac{1}{u + \sqrt{2}} - \frac{1}{u - \sqrt{2}} \right) \, du$. To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1}{\tan t + \sqrt{2}} - \frac{1}{\tan t - \sqrt{2}} \right) \sec^2 t \, dt$.

Simplifying the Integral

To simplify the integral, we can use the trigonometric identity $\sec^2 t = 1 + \tan^2 t$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1 + \tan^2 t}{\tan t + \sqrt{2}} - \frac{1 + \tan^2 t}{\tan t - \sqrt{2}} \right) \, dt$.

Evaluating the Integral

To evaluate this integral, we can use the substitution $v = \tan t$, which gives us $dv = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{\sqrt{2}} \int \left( \frac{1 + v^2}{v + \sqrt{2}} - \frac{1 + v^2}{v - \sqrt{2}} \right) \, dv$.

Simplifying the Integral

To simplify the integral, we can use the method of partial fractions. We can write the integral as $\frac{1}{\sqrt{2}} \int \left( \frac{A}{v + \sqrt{2}} + \frac{B}{v - \sqrt{2}} \right) \, dv$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $1 + v^2 = A(v - \sqrt{2}) + B(v + \sqrt{2})$. Equating the coefficients of $v$, we get $0 = A + B$. Equating the constant terms, we get $1 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the integral as $\frac{1}{2\sqrt{2}} \int \left( \frac{1}{v + \sqrt{2}} - \frac{1}{v - \sqrt{2}} \right) \, dv$. To evaluate this integral, we can use the substitution $v = \tan t$, which gives us $dv = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1}{\tan t + \sqrt{2}} - \frac{1}{\tan t - \sqrt{2}} \right) \sec^2 t \, dt$.

Simplifying the Integral

To simplify the integral, we can use the trigonometric identity $\sec^2 t = 1 + \tan^2 t$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1 + \tan^2 t}{\tan t + \sqrt{2}} - \frac{1 + \tan^2 t}{\tan t - \sqrt{2}} \right) \, dt$.

Evaluating the Integral

To evaluate this integral, we can use the substitution $u = \tan t$, which gives us $du = \sec^2 t \, dt$. Therefore, the integral becomes $\frac{1}{2\sqrt{2}} \int \left( \frac{1 + u^2}{u + \sqrt{2}} - \frac{1 + u^2}{u - \sqrt{2}} \right) \, du$.

Simplifying the Integral

To simplify the integral, we can use the method of partial fractions. We can write the integral as $\frac{1}{2\sqrt{2}} \int \left( \frac{A}{u + \sqrt{2}} + \frac{B}{u - \sqrt{2}} \right) \, du$. To find the values of $A$ and $B$, we can equate the numerator of the original integral to the numerator of the partial fraction decomposition. We have $1 + u^2 = A(u - \sqrt{2}) + B(u + \sqrt{2})$. Equating the coefficients of $u$, we get $0 = A + B$. Equating the constant terms, we get $1 = -A\sqrt{2} + B\sqrt{2}$. Solving these equations, we get $A = \frac{1}{2}$ and $B = -\frac{1}{2}$.

Evaluating the Integral

Now that we have found the values of $A$ and $B$, we can rewrite the integral as $\