1
$\begingroup$

What is the integral of $\int \frac{\cos \left(x\right)}{\sin ^2\left(x\right)+\sin \left(x\right)}dx$ ?

I understand one can substitute $u=\tan \left(\frac{x}{2}\right)$

and one can get (1) $\int \frac{\frac{1-u^2}{1+u^2}}{\left(\frac{2u}{1+u^2}\right)^2+\frac{2u}{1+u^2}}\frac{2}{1+u^2}du$

but somehow this simplifies to (2) $\int \frac{1}{u}-\frac{2}{u+1}du$

to get: (3) $\ln \left(\tan \left(\frac{x}{2}\right)\right)-2\ln \left(\tan \left(\frac{x}{2}\right)+1\right)+C$

as the final answer. But how does one simplify (1) to (2)?

$\endgroup$
2
1
$\begingroup$

Hints; let $u=\sin x$ and complete the square on the new denominator; $u^2+u=(u+\frac12 )^2-\frac14$

It's all trivial substitutions and algebraic manipulations from here.

Weirstrass substitution is too much of a trouble here, I reccomend not going that way.

$\endgroup$
1
$\begingroup$

Consider the integral \begin{align} I = \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } . \end{align}

Method 1


Make the substitution $u = \tan\left(\frac{x}{2}\right)$ for which \begin{align} \cos(x) &= \frac{1-u^{2}}{1+u^{2}} \\ \sin(x) &= \frac{2u}{1+u^{2}} \\ dx &= \frac{2 \, du}{1+u^{2}} \end{align} and the integral becomes \begin{align} I &= \int \frac{ \frac{1-u^{2}}{1+u^{2}} \cdot \frac{2 \, du}{1+u^{2}} }{ \left( \frac{2u}{1+u^{2}} \right)^{2} + \frac{2u}{1+u^{2}} } \\ &= \int \frac{2(1-u^{2})}{(1+u^{2})^{2}} \frac{ (1+u^{2})^{2} }{2u} \, \frac{du}{2u+ 1 + u^{2} } \\ &= \int \frac{(1-u^{2})^{2} \, du}{ u (1+u)^{2} } = \int \frac{(1-u) \, du}{u(1+u)} \\ &= \int \left( \frac{1}{u} - \frac{2}{1+u} \right) \, du\\ &= \ln(u) - 2 \ln(1+u) + c_{1}. \end{align} This leads to \begin{align} \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } = \ln \left( \frac{\tan\left(\frac{x}{2}\right)}{ \left( 1 + \tan\left(\frac{x}{2}\right) \right)^{2} } \right) + c_{1} \end{align}

Method 2


Make the substitution $u = \sin(x)$ to obtain \begin{align} I &= \int \frac{du}{u^{2} + u} = \int \left( \frac{1}{u} - \frac{1}{1+u} \right) \, du \\ &= \ln(u) - \ln(1+u) + c_{2} \end{align} for which \begin{align} \int \frac{\cos(x) \, dx}{\sin^{2}(x) + \sin(x) } = \ln\left( \frac{\sin(x)}{1+\sin(x) } \right) + c_{2} \end{align}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.