1. ## Fraction Integral

Find $\displaystyle \int\frac{dx}{4-5\sin\:x}$

Thanks

2. Originally Posted by polymerase
Find $\displaystyle \int\frac{dx}{4-5sin\:x}$
Let $\displaystyle t = \tan \frac{x}{2}$ then $\displaystyle \sin x =\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}= \frac{1-t^2}{1+t^2}$ and $\displaystyle t' = \frac{1}{2}\sec^2 \frac{x}{2} = \frac{1}{2}\left( 1+\tan^2 \frac{x}{2}\right) =\frac{1}{2}(1+t^2)$.

Thus, it means,
$\displaystyle \int \frac{dx}{4-5\sin x} = \int \frac{dx}{4 - 5 \left( \frac{1-t^2}{1+t^2} \right)} = $$\displaystyle \int \frac{1+t^2 ~ dx}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{9t^2-1} 3. Originally Posted by ThePerfectHacker Let \displaystyle t = \tan \frac{x}{2} then \displaystyle \sin x =\frac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}= \frac{1-t^2}{1+t^2} and \displaystyle t' = \frac{1}{2}\sec^2 \frac{x}{2} = \frac{1}{2}\left( 1+\tan^2 \frac{x}{2}\right) =\frac{1}{2}(1+t^2). Thus, it means, \displaystyle \int \frac{dx}{4-5\sin x} = \int \frac{dx}{4 - 5 \left( \frac{1-t^2}{1+t^2} \right)} =$$\displaystyle \int \frac{1+t^2 ~ dx}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{4(1+t^2) - 5(1-t^2)} = 2\int \frac{dt}{9t^2-1}$
or you could have just let $\displaystyle t=tan\frac{x}{2}$ and then use substitutions of $\displaystyle sin\:x=\frac{2t}{1+t^2}$ and $\displaystyle dx=\frac{2}{1+t^2}$ This leads to $\displaystyle \int\frac{dt}{2t^2-5t+2}$.

4. Originally Posted by polymerase
or you could have just let $\displaystyle t=tan\frac{x}{2}$ and then use substitutions of $\displaystyle sin\:x=\frac{2t}{1+t^2}$ and $\displaystyle dx=\frac{2}{1+t^2}$ This leads to $\displaystyle \int\frac{dt}{2t^2-5t+2}$.
I made a mistake (I was on the bottom of my vodka bottle when I posted this) $\displaystyle \sin x = \frac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$. Now do what I did and you get your answer.