Practicing Success
Practicing Success
CUET Preparation Today
| What is $\int_{0}^{\pi/4}\log(1+\tan x)dx$? |
1 $(\pi/4)\log3$ $(\pi/8)\log 2$ $\pi/2$ |
| $(\pi/8)\log 2$ |
| Using ,$\int_{0}^{a}f(x)dx$=$\int_{0}^{a}f(a-x)dx$ we get $I=\int_{0}^{\pi/4}\log(1+\tan (\pi/4-x))dx=\int_{0}^{\pi/4}\log(2/1+tan x)dx$. From this we can write $2I=[x\log 2](\pi/4)-[x\log 2](0)$. Hence $I=\pi/8\log 2$ |
