Integral evaluation (step-by-step)

Question

I'm trying to evaluate the integral by exponent. Could you help me with following steps?

Integral: $$\int \frac{1}{4+\sin(x)} dx$$

$$\sin x=\frac{e^{ix}-e^{-ix}}{2i}$$

$$\int \frac{1}{4+sin(x)} dx = \int \frac{1}{4+\frac{e^{ix}-e^{-ix}}{2i}} dx = \int \frac{2i}{8i+e^{ix}-e^{-ix}} dx$$

Is it possible? Or any ways exist?

Thanks in advance.

Answer 1

Let be $t = \tan(\frac{x}{2})$ so that $$\sin x=\frac{2t}{1+t^2},\quad\cos x=\frac{1-t^2}{1+t^2},\quad\operatorname{d}\!x=\frac{2 \operatorname{d}\!t}{1 + t^2}$$ and the integral becomes $$ \int \frac{\operatorname{d}\!t}{2t^2+t+2}=\frac{1}{2}\int \frac{\operatorname{d}\!t}{\left(t+\frac{1}{4}\right)^2+\frac{15}{16}}=8\int \frac{\operatorname{d}\!t}{\left(\frac{4t+1}{\sqrt{15}}\right)^2+1} $$ Then putting $u=\frac{4t+1}{\sqrt{15}},\,\operatorname{d}\!u=\frac{4}{\sqrt{15}}\operatorname{d}\!t$ we obtain $$ \frac{2}{\sqrt{15}}\int \frac{\operatorname{d}\!t}{u^2+1}=\frac{2}{\sqrt{15}}\arctan u+\text{constant}. $$ Finally we have $u=\frac{4t+1}{\sqrt{15}}=\frac{4\tan(\frac{x}{2})+1}{\sqrt{15}}$ and the integral is $$ \int\frac{\operatorname{d}\!x}{4+\sin x}=\frac{2}{\sqrt{15}}\arctan\left(\frac{4\tan(\frac{x}{2})+1}{\sqrt{15}}\right)+\text{constant} $$

Answer 2

$$a > \left| b \right|;\frac{{2\pi }}{{\sqrt {a^2 - b^2 } }} = \int\limits_0^{2\pi } {\frac{{d\theta }}{{a + b\sin \theta }}} $$

Answer 3

with $$\sin(x)=\frac{2t}{1+t^2}$$ and $$dx=\frac{2}{1+t^2}dt$$ we get $$\int\frac{1}{2t^2+t+2}dt$$ a rational intagral hint: the result is $$\frac{2 \tan ^{-1}\left(\frac{4 t+1}{\sqrt{15}}\right)}{\sqrt{15}}$$

Answer 4

you can use:

$u = \tan \frac{x}{2} \Leftrightarrow \left\{ \begin{array}{l} \sin x = \frac{{2u}}{{1 + u^2 }} \\ dx = \frac{{2du}}{{1 + u^2 }} \\ x = 2\arctan u \\ \end{array} \right.$

Answer 5

We can rewrite your integral as $$ \int \frac{2i\, e^{ix}}{e^{2ix} + 8i \, e^{ix} - 1}dx $$ We can make the substitution $u = e^{ix}$ to rewrite this as $$ \int \frac{2}{u^2 + 8i\,u - 1}\,du $$ This is now the integral of a rational expression, which can be evaluated using partial frations.

Answer 6

To make it more general, let us consider the case of $$I=\int \frac{dx}{a+b\sin(x)+c\cos(x)}$$ and, as other answers proposed, use the tangent half-angle substitution (Weierstrass substitution) $$t = \tan(\frac{x}{2}),\quad\sin x=\frac{2t}{1+t^2},\quad\cos x=\frac{1-t^2}{1+t^2},\quad\operatorname{d}\!x=\frac{2 \operatorname{d}\!t}{1 + t^2}$$ This will lead to $$I=2\int \frac{dt}{(a-c)t^2+2bt+(a+c)}$$ Completing the square $$(a-c)t^2+2bt+(a+c)=(a-c)\Big((t+\frac{b}{a+c})^2+\frac{a^2-b^2-c^2}{(a-c)^2}\Big)$$ $$I=\frac{2}{a-c}\int \frac{dt}{\left(t+\frac{b}{a-c}\right)^2+\frac{a^2-b^2-c^2}{(a-c)^2}}$$ where you recognize some standard antiderivatives.

Answer 7

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large\int{\dd x \over 4 + \sin\pars{x}}} =\int{4 - \sin\pars{x} \over 16 - \sin^{2}\pars{x}}\,\dd x =\int{4 - \sin\pars{x} \over 15 + \cos^{2}\pars{x}}\,\dd x \\[5mm]&=\int{4 \over 15 + \cos^{2}\pars{x}}\,\dd x -\int{\sin\pars{x} \over 15 + \cos^{2}\pars{x}}\,\dd x \\[5mm]&=4\int{\sec^{2}\pars{x} \over 15\sec^{2}\pars{x} + 1}\,\dd x -\int{\sin\pars{x} \over 15 + \cos^{2}\pars{x}}\,\dd x \\[5mm]&={4 \over \root{15}}\int{\root{15}\sec^{2}\pars{x}/4 \over \bracks{\root{15}\tan\pars{x}/4}^{2} + 1}\,\dd x -{1 \over \root{15}}\int{\sin\pars{x}/\root{15} \over \bracks{\cos\pars{x}/\root{15}}^{2} + 1}\,\dd x \\[1cm]&={4\root{15} \over 15} \int{1 \over \bracks{\root{15}\tan\pars{x}/4}^{2} + 1} \,\dd\bracks{\root{15}\tan\pars{x} \over 4} \\[5mm]&+{\root{15} \over 15}\int{\dd\bracks{\cos\pars{x}/\root{15}} \over \bracks{\cos\pars{x}/\root{15}}^{2} + 1} \\[1cm]&=\color{#66f}{\large{4\root{15} \over 15}\,\arctan\pars{{\root{15} \over 4}\,\tan\pars{x}} +{\root{15} \over 15}\,\arctan\pars{{\root{15} \over 15}\,\cos\pars{x}}} \\[5mm]&+\mbox{a constant} \end{align}