Find the value of the following integral. $$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} + \int_{1/e}^{\cot x} \cfrac{dt}{1+t^2}$$
As a start, I've done this -
$$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} - \int_{\cot x}^{1/e} \cfrac{dt}{1+t^2} \\
=\int_{\cot x}^{\tan x} \cfrac{t-1}{1+t^2} dt $$
Is this valid? If yes, then how to proceed?
Answer
$$\int_{1/e}^{\tan x} \cfrac{tdt}{1+t^2} + \int_{1/e}^{\cot x} \cfrac{dt}{1+t^2}$$
For the first integral, substitute $1+t^2=s\implies tdt= \dfrac {ds}2$. Also, we know that $\int\frac 1{1+t^2} dt =\arctan t +c$. We thus get: $$\int_{1/e}^{\tan x} \dfrac{ds}{2s} + \arctan (\cot x)-\arctan\left (\dfrac 1e\right )$$ $$=\dfrac 12(\ln \tan x +1)+\arctan (\cot x)-\arctan\left (\dfrac 1e\right )$$
No comments:
Post a Comment