Question

Question #03e6a

Answer 1

`=1/2 (tan^(-1) x+x/(1+x^2))`

Explanation:

`int_0^(tan^(-1) x)1/(1+tan^2 t) dt`

By Trigonometric Identities,

#=int_0^(tan^(-1) x)1/sec^2 t dt =int_0^(tan^(-1) x)cos^2 t dt =int_0^(tan^(-1)x)1/2(1+cos 2t)dt#

By integrating and `sin 2t=2sin t cos t`

#=1/2 [1+(sin 2t)/2]_0^(tan^(-1) x) =1/2[t+sin t cos t]_0^(tan^(-1) x)#

`=1/2(tan^(-1) x +sin(tan^(-1) x) cdot cos(tan^(-1) x))`

`=1/2(tan^(-1) x+ x/sqrt(1+x^2) cdot 1/sqrt(1+x^2))`

`=1/2(tan^(-1) x+x/(1+x^2))`

I hope that this was clear.