Question

1∫ dx/eˣ+e⁻ˣ ,Evaluate it.0

Answer 1

we have to evaluate , $\int\limits^1_0{\frac{1}{(e^x+e^{-x})}}\,dx$

let e^x = y ⇒e^-x = 1/y

now differentiate e^-x = y with respect to x,

i.e., e^x dx = dy

or, dx = dy/y [ as e^x = y ]

upper limit : e¹ = e

lower limit : e^0 = 1

now,$\int\limits^1_0{\frac{1}{(e^x+e^{-x})}}\,dx$ comverts into $\int\limits^e_1{\frac{1}{(y+1/y)}}\,dy/y$

= $\int\limits^e_1{\frac{1}{(y^2+1)}}\,dy$

= $\left[tan^{-1}y\right]^e_1$

= tan-¹(e) - tan-¹(1)

= tan-¹(e) - π/4

Answer 2

Answer:

The result is:

$tan^{-1}(e)-\dfrac{\pi}{4}$

Step-by-step explanation:

Given equation is:

$\displaystyle\int _0^1\dfrac{dx}{e^x+e^{-x}}$

We may write the equation as:

$\displaystyle\int _1^2\dfrac{dx}{e^x+\frac{1}{e^{x}}}$

Let:

$e^x=z$

Differentiating with respect to x.

$e^x\,dx=dz$

Now substituting for the value of limit:

When $x=0$

$z=e^0\\z=1$

When $x=1$

$z=e^1\\z=e$

Therefore integrating by substituting the values:

$\displaystyle\int _1^e\dfrac{1}{z+\frac{1}{z}}\cdot \dfrac{dz}{z}\\=\displaystyle\int _1^e\dfrac{1}{\dfrac{z^2+1}{z}}\cdot \dfrac{dz}{z}\\=\displaystyle\int _1^e\dfrac{1}{z^2+1}dz\\=[tan^{-1}(z)]_1^e$

Substituting the values of limit in the equation:

$[tan^{-1}(e)-tan^{-1}(1)]\\=tan^{-1}(e)-\dfrac{\pi}{4}$