Let f(x) = intlimits_1^x frac{tan^{-1} t}{t} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We are given $f(x) = \int_{1}^{x} \frac{	an^{-1} t}{t} dt$. We need to evaluate $f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an

Vedclass · Accepted Answer

$\pi$

Vedclass · Answer

(B) We are given $f(x) = \int_{1}^{x} \frac{	an^{-1} t}{t} dt$. We need to evaluate $f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an^{-1} t}{t} dt - \int_{1}^{1/e^2} \frac{	an^{-1} t}{t} dt$. Note that $-\int_{1}^{1/e^2} \frac{	an^{-1} t}{t} dt = \int_{1/e^2}^{1} \frac{	an^{-1} t}{t} dt$. So,$f(e^2) - f\left(\frac{1}{e^2}ight) = \int_{1}^{e^2} \frac{	an^{-1} t}{t} dt + \int_{1/e^2}^{1} \frac{	an^{-1} t}{t} dt = \int_{1/e^2}^{e^2} \frac{	an^{-1} t}{t} dt$. Let $I = \int_{1/e^2}^{e^2} \frac{	an^{-1} t}{t} dt$. Using the property $\int_{a}^{b} g(t) dt = \int_{a}^{b} g\left(\frac{ab}{t}ight) dt$,where $ab = (e^2)(1/e^2) = 1$,we have $I = \int_{1/e^2}^{e^2} \frac{	an^{-1}(1/t)}{t} dt$. Adding the two expressions for $I$: $2I = \int_{1/e^2}^{e^2} \frac{	an^{-1} t + 	an^{-1}(1/t)}{t} dt$. Since $	an^{-1} t + 	an^{-1}(1/t) = \frac{\pi}{2}$ for $t > 0$,we get $2I = \int_{1/e^2}^{e^2} \frac{\pi/2}{t} dt = \frac{\pi}{2} [\ln t]_{1/e^2}^{e^2} = \frac{\pi}{2} (\ln e^2 - \ln e^{-2}) = \frac{\pi}{2} (2 - (-2)) = \frac{\pi}{2} (4) = 2\pi$. Therefore,$I = \frac{2\pi}{2} = \pi$.

Let $f(x) = \int\limits_1^x \frac{\tan^{-1} t}{t} dt$ for $x > 0$. Then $f(e^2) - f\left(\frac{1}{e^2}\right)$ is

Similar Questions

Evaluate the definite integral: $\int_{\frac{1}{2}}^{2} \frac{x^2 \ln x}{(1+x^2)^3} dx$

The value of $\int_{0}^{\pi} x \sin^{3} x \, dx$ is

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

The value of the integral $\sum\limits_{k = 1}^n {\int_0^1 {f(k - 1 + x)\,dx} } $ is

$\int_{0}^{\infty} \frac{x \ln x}{(1 + x^2)^2} \, dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module