int_(1//e)^(tan x)(tdt)/(1+t^(2))+int_(1//e)^(cot x)(dt)/(t(1+t^(2)))

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Evaluate First Integral: Let's start by evaluating the first integral: $$ \int_{1/e}^{\tan(x)} \frac{t}{1+t^2} dt $$ To solve this integral, we will use the substitution method. Let $ u = 1 + t^2 $, then $ du = 2t dt $. We need to adjust the differential, so we have $ \frac{1}{2} du = t dt $.Substitution Method: Now, we substitute $ u $ and $ \frac{1}{2} du $ into the integral: $$ \int_{1/e}^{\tan(x)} \frac{t}{1+t^2} dt = \frac{1}{2} \int_{u(1/e)}^{u(\tan(x))} \frac{1}{u} du $$ We need to evaluate the new limits of integration by substituting $ t = 1/e $ and $ t = \tan(x) $ into $ u = 1 + t^2 $.Calculate New Limits: Calculate the new limits of integration: When $ t = 1/e $, $ u = 1 + (1/e)^2 $. When $ t = \tan(x) $, $ u = 1 + \tan^2(x) $. Note that $ 1 + \tan^2(x) $ is equal to $ \sec^2(x) $ by the Pythagorean identity. So the new limits are $ u(1/e) = 1 + (1/e)^2 $ and $ u(\tan(x)) = \sec^2(x) $.Integrate with Respect to u: Now we can integrate with respect to $ u $: $$ \frac{1}{2} \int_{1 + (1/e)^2}^{\sec^2(x)} \frac{1}{u} du = \frac{1}{2} \left[ \ln|u| \right]_{1 + (1/e)^2}^{\sec^2(x)} $$Evaluate Definite Integral: Evaluate the definite integral: $$ \frac{1}{2} \left[ \ln|\sec^2(x)| - \ln|1 + (1/e)^2| \right] $$ $$ = \frac{1}{2} \ln|\sec^2(x)| - \frac{1}{2} \ln|1 + (1/e)^2| $$Evaluate Second Integral: Now let's evaluate the second integral: $$ \int_{1/e}^{\cot(x)} \frac{dt}{t(1+t^2)} $$ For this integral, we will use the substitution method again. Let $ v = 1 + t^2 $, then $ dv = 2t dt $. We need to adjust the differential, so we have $ \frac{1}{2} dv = t dt $.Substitution Method: Substitute $ v $ and $ \frac{1}{2} dv $ into the integral: $$ \int_{1/e}^{\cot(x)} \frac{dt}{t(1+t^2)} = \frac{1}{2} \int_{v(1/e)}^{v(\cot(x))} \frac{1}{v} dv $$ We need to evaluate the new limits of integration by substituting $ t = 1/e $ and $ t = \cot(x) $ into $ v = 1 + t^2 $.Calculate New Limits: Calculate the new limits of integration for the second integral: When $ t = 1/e $, $ v = 1 + (1/e)^2 $. When $ t = \cot(x) $, $ v = 1 + \cot^2(x) $. Note that $ 1 + \cot^2(x) $ is equal to $ \csc^2(x) $ by the Pythagorean identity. So the new limits are $ v(1/e) = 1 + (1/e)^2 $ and $ v(\cot(x)) = \csc^2(x) $.Integrate with Respect to v: Now we can integrate with respect to $ v $: $$ \frac{1}{2} \int_{1 + (1/e)^2}^{\csc^2(x)} \frac{1}{v} dv = \frac{1}{2} \left[ \ln|v| \right]_{1 + (1/e)^2}^{\csc^2(x)} $$Evaluate Definite Integral: Evaluate the definite integral: $$ \frac{1}{2} \left[ \ln|\csc^2(x)| - \ln|1 + (1/e)^2| \right] $$ $$ = \frac{1}{2} \ln|\csc^2(x)| - \frac{1}{2} \ln|1 + (1/e)^2| $$Combine Results: Combine the results from Step 5 and Step 10: $$ \frac{1}{2} \ln|\sec^2(x)| - \frac{1}{2} \ln|1 + (1/e)^2| + \frac{1}{2} \ln|\csc^2(x)| - \frac{1}{2} \ln|1 + (1/e)^2| $$ $$ = \frac{1}{2} \ln|\sec^2(x)| + \frac{1}{2} \ln|\csc^2(x)| - \ln|1 + (1/e)^2| $$Simplify Final Expression: Simplify the final expression: Since $ \ln|a| + \ln|b| = \ln|ab| $, we can combine the logarithms: $$ \frac{1}{2} \ln|\sec^2(x)\csc^2(x)| - \ln|1 + (1/e)^2| $$ $$ = \frac{1}{2} \ln|\sec^2(x)\csc^2(x)| - \ln|1 + (1/e)^2| $$ Note that $ \sec^2(x)\csc^2(x) = \frac{1}{\cos^2(x)}\frac{1}{\sin^2(x)} = \frac{1}{\sin^2(x)\cos^2(x)} = \frac{1}{\sin^2(x)\cos^2(x)} $.Recognize that $ \sin^2(x)\cos^2(x) $ is the square of $ \sin(x)\cos(x) $, which is $ \frac{1}{4}\sin^2(2x) $ by the double-angle formula. Therefore, we have: $$ \frac{1}{2} \ln|\frac{1}{\frac{1}{4}\sin^2(2x)}| - \ln|1 + (1/e)^2| $$ $$ = \frac{1}{2} \ln|4\csc^2(2x)| - \ln|1 + (1/e)^2| $$The final simplified answer is: $$ \frac{1}{2} \ln|4\csc^2(2x)| - \ln|1 + (1/e)^2| $$

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

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