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Q.
The area of the region above the x-axis bounded by the curve $y = \tan \, x, 0 \le x \le \frac{\pi}{2}$ and the tangent to the curve at $x = \frac{\pi}{4}$ is :
Application of Integrals
Solution:
The given curve is $y = \tan x$ ...(1)
when $x = \frac{\pi}{4} , y = 1$
Equation of tangent at P is
$ y -1 = \left(\sec^{2} \frac{\pi}{4}\right) \left(x - \frac{\pi}{4}\right)$
or $ y = 2x+1 - \frac{\pi}{2} $ ...(2)
Area of shaded region
= area of OPMO - ar ($\Delta $PLM)
$= \int\limits^{\frac{\pi}{4}}_{0} \tan \,x \,dx - \frac{1}{2} \left(OM -OL\right)PM $
$= \left[\log \sec x\right]^{\frac{\pi}{4}}_{0} - \frac{1}{2} \left\{\frac{\pi}{4} - \frac{\pi-2}{4}\right\} \times 1$
$ = \frac{1}{2} \left[\log \, 2 - \frac{1}{2} \right]$ sq unit
