# Indefinite Integral

#### jacks

##### Well-known member
$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx$, where $a>b$

My Trial :: Using Integration by parts::

$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx = \frac{1}{b}\int -\csc (x)\cdot \frac{-b\sin x}{(a+b\sin x)^2}dx$

$\displaystyle -\frac{1}{b}\cdot \csc (x)\cdot \frac{-1}{(a+b\sin x)}+\int (-\csc x \cdot \cot x)\cdot \frac{-1}{(a+b\sin x)}dx$

Now How Can I Calculate (II) Integral

Help me

Thanks

#### chisigma

##### Well-known member
$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx$, where $a>b$

My Trial :: Using Integration by parts::

$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx = \frac{1}{b}\int -\csc (x)\cdot \frac{-b\sin x}{(a+b\sin x)^2}dx$

$\displaystyle -\frac{1}{b}\cdot \csc (x)\cdot \frac{-1}{(a+b\sin x)}+\int (-\csc x \cdot \cot x)\cdot \frac{-1}{(a+b\sin x)}dx$

Now How Can I Calculate (II) Integral

Help me

Thanks
For integrals of this type the following substitution is often useful...

$\displaystyle t = \tan \frac{x}{2}$

$\displaystyle dx = \frac{2}{1+ t^{2}}\ d t$

$\displaystyle \sin x = \frac{2\ t}{1 + t^{2}}$

$\displaystyle \cos x = \frac{1 - t^{2}}{1 + t^{2}}$

Kind regards

$\chi$ $\sigma$

#### jacks

##### Well-known member
Thanks chisigma for giving me a hint.

But Using this method, calculation of integral is very complex .

Can anyone have a better method.

If yes the please explain here

Thanks

#### chisigma

##### Well-known member
$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx$, where $a>b$

My Trial :: Using Integration by parts::

$\displaystyle \int\frac{1}{(a+b\sin x)^2}dx = \frac{1}{b}\int -\csc (x)\cdot \frac{-b\sin x}{(a+b\sin x)^2}dx$

$\displaystyle -\frac{1}{b}\cdot \csc (x)\cdot \frac{-1}{(a+b\sin x)}+\int (-\csc x \cdot \cot x)\cdot \frac{-1}{(a+b\sin x)}dx$

Now How Can I Calculate (II) Integral

Help me

Thanks
In the Italian translation of M.H. Spiegel Mathematic Handbook on page 76 is written...

$\displaystyle \int \frac{d x}{(a + b\ \sin x)^{2}} = \frac{b\ \cos x}{a\ (a^{2} - b^{2})\ (a + b\ \sin x)} + \frac{2\ b}{(a^{2}-b^{2})^{\frac{3}{2}}}\ \tan^{-1} \frac{a\ \tan \frac{x}{2} + b}{\sqrt{a^{2}- b^{2}}} + c\ (1)$

... where [of course...] $a^{2} \ne b^{2}$...

Kind regards

$\chi$ $\sigma$

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