Find $\int_{1/e}^{\tan x}\frac{dt}{1+t^2}+\int_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$

Question

Evaluate $$\int_{1/e}^{\tan x}\frac{dt}{1+t^2}+\int_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$$

Using Leibniz rule $$ \frac{d}{dt}\int_{a(t)}^{b{t}}f(x)dx=f(b(t))\frac{d}{dt}b(t)-f(a(t))\frac{d}{dt}a(t) $$ Using that $$ I=I_1+I_2=\int_{1/e}^{\tan x}\frac{dt}{1+t^2}+\int_{1/e}^{\cot x}\frac{dt}{t(1+t^2)} $$ $$ I^{'}_1(x)=\frac{\sec^2x}{1+\tan^2x}=1\implies I_1(x)=x+C_1\\ I_1(0)=C_1=\Big[\tan^{-1}x\Big]_{1/e}^0=-\tan^{-1}(1/e)\implies \boxed{I_1(x)=x-\tan^{-1}(1/e)} $$ $$ I^{'}_2(x)=\frac{-\csc^2x}{\cot x(1+\cot^2x)}=-\tan x\implies I_2(x)=-\log|\sec x|+C_2=\log|\cos x|+C\\ I_2(\pi/4)=-\log\sqrt{2}+C_2=-\frac{1}{2}\log 2+C_2=\int_{1/e}^{1}\Big[\frac{1}{t}-\frac{t}{1+t^2}\Big]dt=\Big[\log t-\frac{1}{2}\log|1+t^2|\Big]_{1/e}^1\\ -\frac{1}{2}\log 2+C_2=-\frac{1}{2}\log 2+1-\frac{1}{2}\log|1+\frac{1}{e^2}|\\ C_2=1-\frac{1}{2}\log|1+\frac{1}{e^2}| $$ If $1$ is the solution taking the derivating of $I(t)$ is supposed to give zero right, so what is going wrong here ?

Note: The solution given in my reference is $1$

From where did you get this problem? And did you copy it correctly? — Zacky, Jan 11 '20 at 21:20
Does this answer your question? Evaluation of $\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$. If the result is $1$ then the first integral should be $\frac{t}{1+t^2}$ and you can find more in that link. — Zacky, Jan 11 '20 at 21:38
If the answer is supposed to be $1$, then you may have a typo in the problem. See the link in the Zacky's comment — bjorn93, Jan 11 '20 at 21:47
@Zacky I hate this kinda typo in books, a simple letter can make the whole thing different. — Sooraj S, Jan 11 '20 at 21:57

score 1 · Answer 1 · answered Jan 11 '20 at 21:22

In order that both expressions are defined over an interval, we have to assume $0<x<\pi/2$.

Note that $$ \int_{1/e}^{\tan x}\frac{1}{1+t^2}\,dt=\Bigl[\arctan t\Bigr]_{1/e}^{\tan x}=x-\arctan\frac{1}{e} $$ For the second integral, the partial fraction decomposition is $$ \frac{1}{t(1+t^2)}=\frac{1}{t}-\frac{t}{1+t^2} $$ so we have $$ \int_{1/e}^{\cot x}\frac{1}{t(1+t^2)}\,dt=\Bigl[\log t-\frac{1}{2}\log(1+t^2)\Bigr]_{1/e}^{\cot x} $$ The expression with $t=\cot x$ is $$ \log\cot x-\frac{1}{2}\log(1+\cot^2x)=\log\cos x-\log\sin x+\log\sin x=\log\cos x $$ For $t=1/e$ we get $$ -1-\frac{1}{2}\log(e^2+1)+1=-\log\sqrt{e^2+1} $$ Thus the final answer is $$ x-\arctan\frac{1}{e}+\log\cos x+\log\sqrt{e^2+1} $$

If the problem is $$ \int_{1/e}^{\tan x}\frac{1}{1+t^2}\,dt+\int_{1/e}^{\cot x}\frac{1}{1+t^2}\,dt $$ then the second integral is $$ \arctan\cot x-\arctan\frac{1}{e}=\frac{\pi}{2}-x-\arctan\frac{1}{e} $$ and the whole expression is $$ \frac{\pi}{2}-2\arctan\frac{1}{e} $$

Find $\int_{1/e}^{\tan x}\frac{dt}{1+t^2}+\int_{1/e}^{\cot x}\frac{dt}{t(1+t^2)}$

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