Question

appropriate answers needed!

$\displaystyle\sf\int\limits_{1/e}^{tanx} \dfrac{tdt}{1+t^2} + \int\limits_{1/e}^{cotx} \dfrac{dt}{t(1+t^2)}$​

Answer 1

Question:

Evaluate the Integral :

$\sf\int\limits_{1/e}^{tanx} \dfrac{tdt}{1+t^2}+\int\limits_{1/e}^{cotx}\dfrac{dt}{t(1+t^2)}$

Solution :

We have to find the integral of :

$\sf\int\limits_{1/e}^{tan(x)} \dfrac{tdt}{1+t^2}+\int\limits_{1/e}^{cot(x)}\dfrac{dt}{t(1+t^2)}$

Let $\sf\int\limits_{1/e}^{tan(x)}\dfrac{tdt}{1+t^2}=I_1$

and $\sf\int\limits_{1/e}^{cot(x)}\dfrac{dt}{t(1+t^2)}=I_2$

Let's First solve $\sf\:I_1$

$\sf\:I_1=\int\limits_{1/e}^{tan(x)}\dfrac{tdt}{1+t^2}$

Let , $\sf\:1+t^2=p$

Now differentiate it with respect to t

$\sf\implies\dfrac{dp}{dt}=2t$

$\sf\implies\:dt=\dfrac{dp}{2t}$

Now , Solve the integrate , same as indefinite integral .

$\int\dfrac{tdt}{1+t^2}$

$\sf=\int\dfrac{t}{p}\times\dfrac{dp}{2t}$

$\sf=\dfrac{1}{2}\dfrac{dp}{p}$

$\sf=\dfrac{1}{2}[\log(p)]$

$\sf=\dfrac{1}{2}[\log(1+t^2)]+c$

Now Apply the Limits , Then ,

$\sf\int\limits_{1/e}^{tan(x)}\dfrac{tdt}{1+t^2}$

$\sf=\dfrac{\log(tan^2x+1}{2}-\dfrac{\log[e^{-2}(e^2+1)]}{2}$

$\sf=\log[sec(x)]-\dfrac{\log(e^{-2}+1)}{2}$

Now , let's solve , $\int\limits_{1/e}^{cot(x)}\dfrac{dt}{t(1+t^2)}=I_2$

Firstly , We have solve it as indefinite, then after that we will apply limits

$\sf\int\dfrac{dt}{t(1+t^2)}$

$\sf=\int\dfrac{(1+t^2-t^2)dt}{t(1+t^2)}$

$\sf=\int[\dfrac{(1+t^2)dt}{t(1+t^2)}-\dfrac{(t^2)dt}{t(1+t^2)}]$

$\sf=\int[\dfrac{dt}{t}-\dfrac{(t)dt}{1+t^2}]$

$\sf=\log(t)-\dfrac{1}{2}[\log(1+t^2)]+c$

Now Apply the Limits, Then

$\sf\int\limits_{1/e}^{cot(x)}\dfrac{dt}{t(1+t^2)}$

$\sf=-\log[\csc(x)]+\log[cot(x)]+\dfrac{\log(e^{-2}+1)}{2}+1$

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Integration by substitution :

The given integral $\sf\int\:f(x)dx$ can be transformed into another by changing the independent variable x to t by substituting x = g(t)