Question

1∫ tan⁻¹x dx ,Evaluate it.0

Answer 1

Answer:

plzz give me brainliest ans and thanks and follow me

Answer 2

Answer:

The answer is:

$\dfrac{\pi}{4} -\dfrac{1}{2}log(2)$.

Step-by-step explanation:

$\displaystyle\int _0^1tan^{-1}x\,dx$

Here we are applying integration by parts formula:

$\displaystyle\int (uv)dx=u\displaystyle\int v\,dx-\displaystyle\int[\dfrac{d}{dx}(u)\displaystyle\int v\,dx]dx$

Therefore:

$=tan^{-1}x\displaystyle\int _0^1dx-\displaystyle\int _0^1[\dfrac{d}{dx}(tan^{-1}x)\displaystyle\int _0^1dx]dx$

Integration of:

$tan^{-1}x=\dfrac{1}{1+x^2}$.

$=[tan^{-1}x\cdot x]_0^1-\displaystyle\int _0^1[\dfrac{1}{1+x^2}\cdot x]dx$

Substituting:

$1+x^2=z\\2x\,dx=dz\\x\,dx=\dfrac{dz}{2}$

Substituting the value of limits:

Lower limit:

When $x=0$

$z=1+0\\z=1$

Upper limit:

When $x=1$

$z=1+1\\z=2$

Hence integrating by substituting the values:

$=[x\,tan^{-1}x]_0^1-\displaystyle\int _1^2[\dfrac{1}{z}\cdot \dfrac{dz}{2}]\\=[x\,tan^{-1}x]_0^1-\dfrac{1}{2}[log(z)]_1^2\\=[1tan^{-1}(1)-0]-\dfrac{1}{2}[log(2)-log(1)]\\=\dfrac{\pi}{4}-\dfrac{1}{2}log(2)$