# Evaluate $\displaystyle \int{\dfrac{1}{1+2\sin{x}} \,} dx$

The trigonometric sine function $\sin{x}$ formed a rational expression in the calculus. The indefinite integration of the trigonometric function has to evaluate with respect to $x$ in this indefinite integration problem.

$\displaystyle \int{\dfrac{1}{1+2\sin{x}} \,} dx$

### Expand the trigonometric function sine

The sine function can be expanded in terms of tan function as per half angle identities.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{1+2 \Bigg(\dfrac{2\tan{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}\Bigg)} \,} dx$

### Simplify the rational trigonometric expression

The trigonometric expression in rational form has to simplify for evaluating the indefinite integration mathematically.

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{1+\dfrac{2 \times 2\tan{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{1+\dfrac{4\tan{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\dfrac{1 \times \Bigg(1+\tan^2{\Big(\dfrac{x}{2}\Big)}\Bigg)+ 4\tan{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1}{\dfrac{1+\tan^2{\Big(\dfrac{x}{2}\Big)}+ 4\tan{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}{1+\tan^2{\Big(\dfrac{x}{2}\Big)}+ 4\tan{\Big(\dfrac{x}{2}\Big)}} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{1+\tan^2{\Big(\dfrac{x}{2}\Big)}}{\tan^2{\Big(\dfrac{x}{2}\Big)}+ 4\tan{\Big(\dfrac{x}{2}\Big)}+1} \,} dx$

$= \,\,\,$ $\displaystyle \int{\dfrac{\sec^2{\Big(\dfrac{x}{2}\Big)}}{\tan^2{\Big(\dfrac{x}{2}\Big)}+ 4\tan{\Big(\dfrac{x}{2}\Big)}+1} \,} dx$

### Differentiate the composite function

In the rational expression, the tan and secant functions are composite functions. Similarly, the differentiation of the tan function is the square of secant function. So, let’s differentiate the tan function by the chain rule.

Take $u = \tan{\Big(\dfrac{x}{2}\Big)}$ and find the differentiation of the composite function by chain rule.

$\implies$ $\dfrac{d}{dx}{\, (u)} \,=\, \dfrac{d}{dx}{\, \tan{\Big(\dfrac{x}{2}\Big)}}$

$\implies$ $\dfrac{du}{dx} \,=\, \sec^2{\Big(\dfrac{x}{2}\Big)} \times \dfrac{d}{dx}{\,\Big(\dfrac{x}{2}\Big)}$

$\implies$ $\dfrac{du}{dx} \,=\, \sec^2{\Big(\dfrac{x}{2}\Big)} \times \dfrac{d}{dx}{\,\Big(\dfrac{1}{2} \times x\Big)}$

$\implies$ $\dfrac{du}{dx} \,=\, \sec^2{\Big(\dfrac{x}{2}\Big)} \times \dfrac{1}{2} \times \dfrac{d}{dx}{\,(x)}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{2} \times \sec^2{\Big(\dfrac{x}{2}\Big)} \times \dfrac{dx}{dx}$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{2} \times \sec^2{\Big(\dfrac{x}{2}\Big)} \times 1$

$\implies$ $\dfrac{du}{dx} \,=\, \dfrac{1}{2} \times \sec^2{\Big(\dfrac{x}{2}\Big)}$

$\implies$ $2 \times du \,=\, 1 \times \sec^2{\Big(\dfrac{x}{2}\Big)} \times dx$

$\implies$ $2du \,=\, \sec^2{\Big(\dfrac{x}{2}\Big)}dx$

Now, transform the rational trigonometric function from $x$ into $u$ by the $u = \tan{\Big(\dfrac{x}{2}\Big)}$ and $2du \,=\, \sec^2{\Big(\dfrac{x}{2}\Big)}dx$.

$\implies$ $\displaystyle \int{\dfrac{\sec^2{\Big(\dfrac{x}{2}\Big)}}{\tan^2{\Big(\dfrac{x}{2}\Big)}+ 4\tan{\Big(\dfrac{x}{2}\Big)}+1} \,} dx$ $\,=\,$ $\displaystyle \int{\dfrac{2 \,du}{u^2+4u+1}}$

### Simplify the Rational algebraic function

The rational expression is converted into algebraic form from trigonometric form. Now, simplify the quadratic expression in the denominator.

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{u^2+4u+1}\,du}$

The quadratic expression cannot be factored but it can be expressed in difference of squares.

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{u^2+2 \times 2 \times u+1}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{u^2+2 \times 2 \times u+1+4-4}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{u^2+2 \times 2 \times u+1+2^2-4}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{u^2+2 \times 2 \times u+2^2+1-4}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{(u+2)^2-3}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2}{(u+2)^2-(\sqrt{3})^2}\,du}$

$=\,\,\,$ $\displaystyle \int{\dfrac{2 \times 1}{(u+2)^2-(\sqrt{3})^2}\,du}$

$=\,\,\,$ $\displaystyle 2 \times \int{\dfrac{1}{(u+2)^2-(\sqrt{3})^2}\,du}$

Now, take $y = u+2$ and differentiate the equation with respect to $u$.

$\implies$ $\dfrac{dy}{du} \,=\, \dfrac{d}{du}{\, (u+2)}$

$\implies$ $\dfrac{dy}{du} \,=\, \dfrac{d}{du}{\, (u)}+\dfrac{d}{du}{\, (2)}$

$\implies$ $\dfrac{dy}{du} \,=\, 1+0$

$\implies$ $\dfrac{dy}{du} \,=\, 1$

$\implies$ $dy \,=\, 1 \times du$

$\implies$ $dy \,=\, du$

$\implies$ $du \,=\, dy$

Now, we can write the integral of the rational expression in terms of $y$.

$\implies$ $\displaystyle 2 \times \int{\dfrac{1}{(u+2)^2-(\sqrt{3})^2}\,du}$ $\,=\,$ $\displaystyle 2 \times \int{\dfrac{1}{y^2-(\sqrt{3})^2}\,dy}$

### Evaluate the Integral of the rational function

The denominator in the rational expression is the difference of the squares of the terms.

$= \,\,\,$ $\displaystyle 2 \times \int{\dfrac{1}{y^2-(\sqrt{3})^2}\,dy}$

The indefinite integration of the multiplicative inverse of the difference of the squares can be evaluated by the integral rule of reciprocal of difference of the squares.

$= \,\,\,$ $2 \times \Bigg(\dfrac{1}{2 \times \sqrt{3}}\log_e{\Bigg|\dfrac{y-\sqrt{3}}{y+\sqrt{3}}\Bigg|}+c_1\Bigg)$

$= \,\,\,$ $\dfrac{2 \times 1}{2 \times \sqrt{3}}\log_e{\Bigg|\dfrac{y-\sqrt{3}}{y+\sqrt{3}}\Bigg|}+2 \times c_1$

$= \,\,\,$ $\require{cancel} \dfrac{\cancel{2} \times 1}{\cancel{2} \times \sqrt{3}}\log_e{\Bigg|\dfrac{y-\sqrt{3}}{y+\sqrt{3}}\Bigg|}+2c_1$

$= \,\,\,$ $\dfrac{1 \times 1}{1 \times \sqrt{3}}\log_e{\Bigg|\dfrac{y-\sqrt{3}}{y+\sqrt{3}}\Bigg|}+c$

$= \,\,\,$ $\dfrac{1}{\sqrt{3}}\log_e{\Bigg|\dfrac{y-\sqrt{3}}{y+\sqrt{3}}\Bigg|}+c$

In this expression, the value of $y$ is $u+2$. So, we can replace it.

$= \,\,\,$ $\dfrac{1}{\sqrt{3}}\log_e{\Bigg|\dfrac{u+2-\sqrt{3}}{u+2+\sqrt{3}}\Bigg|}+c$

Similarly, we have taken that $u = \tan{\Big(\dfrac{x}{2}\Big)}$. So, we can replace the value of $u$ to express the solution in terms of variable $x$.

$= \,\,\,$ $\dfrac{1}{\sqrt{3}}\log_e{\Bigg|\dfrac{\tan{\Big(\dfrac{x}{2}\Big)}+2-\sqrt{3}}{\tan{\Big(\dfrac{x}{2}\Big)}+2+\sqrt{3}}\Bigg|}+c$

Latest Math Topics
Email subscription
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
###### Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.