How would you solve the following integral?
$$\int \frac{1}{(1+x\tan(x))^2} dx$$
Any help would be appreciated.
Asked
Active
Viewed 2,037 times
6
-
I just am not able to get that first substitution. – user34304 Aug 27 '14 at 12:27
-
Do you have some reason to think there's a closed-form solution? The much simpler-looking $\int x\tan x,dx$ doesn't. – Gerry Myerson Aug 27 '14 at 12:36
-
It came in my test. So, I guess it should be a high school level problem. Unless, they made a mistake :o. xD – user34304 Aug 27 '14 at 13:21
2 Answers
13
A substitution works... $$ \frac{1}{(1+x\tan x)^2} = \frac{\cot^2 x}{(\cot x+x)^2} $$ and $$ \frac{d}{dx}\;(\cot x+x) = -\cot^2 x $$ so ... you finish it.
GEdgar
- 111,679
2
$\bf{My\; Solution::}$ Let $\displaystyle I = \int\frac{1}{(1+x\tan x)^2}dx = \int\frac{\cos ^2x}{(x\sin x+\cos x)^2}dx$
So $\displaystyle I = \int \frac{\cos x}{x}\cdot \left\{\frac{x\cos x}{(x\sin x+\cos x)^2}\right\}dx$
Now Using Integration by Parts....
So $\displaystyle I = -\frac{\cos x}{x}\cdot \frac{1}{(x\sin x+\cos x)}-\int \frac{(x\sin x+\cos x)}{x^2}\cdot \frac{1}{(x\sin x+\cos x)}dx$
So $\displaystyle I = -\frac{1}{x\cdot (1+x\tan x)}+\frac{1}{x}+\mathcal{C} = \frac{\tan x}{(1+x\tan x)}+\mathcal{C}.$
juantheron
- 53,015