Area of the region bounded by two curves

Question

Let $y=\tan (x) $ for $0\leq x\leq \dfrac{\pi}{2} $ and a tangent be drawn at $x=\dfrac {\pi}{4} $ then area of eegion bounded above $x$-axis is?

Attempts

Let the equation of tangents be $y-1=m\left( x-\frac{\pi}{4} \right)$ then slope is $\dfrac{dy}{dx}=\sec^2 (x)=2$ at given point.

Thus equation becomes $y=2x-\dfrac{\pi}{2}+1$. Now its typical $f_1(x)-f_2(x) $.

After integrating and subtraction I get answer as $$-\dfrac{\pi^2}{16}+\dfrac{\pi}{4}-\frac{1}{2}\log (2)$$

but the answer given is $$\frac{1}{2}\left( \log (2)-\frac{1}{2} \right)$$

Where is my fault? Thanks!

Answer 1

The tangent line $\;y=2x-\cfrac\pi2+1\;$ crosses the $\;x\,-$ axis at $\;\cfrac\pi4-\frac12\;$ , so the area you want is

$$\int_0^{\frac\pi4-\frac12}\tan x\,dx+\int_{\frac\pi4-\frac12}^{\frac\pi4}\left(\tan x-2x+\frac\pi2-1\right)dx=$$

$$=\left.-\log\cos x\right|_0^{\frac\pi4}-\left.\left(x^2-\left(\frac\pi2-1\right)x\right)\right|_{\frac\pi4-\frac12}^{\frac\pi4}=$$

$$=-\log\cos\frac\pi4+\log\cos0-\left(\frac{\pi^2}{16}-\frac{\pi^2}8+\frac\pi4-\frac{\pi^2}{16}+\frac\pi4-\frac14+\frac{\pi^2}8-\frac\pi2+\frac12\right)=$$

$$=\frac12\log2+\frac14$$

There's a difference of a sign...

Answer 2

I am getting your correct answer with $$\int_{0}^{\frac{1}{4}(-2+\pi)}\tan{x}dx+\int_{\frac{1}{4}(-2+\pi)}^{\frac{\pi}{4}}\tan{x}-(2x-\tfrac{\pi}{2}+1)dx$$ where $\frac{1}{4}(-2+\pi)$ is where the tangent crosses the $x$ axis.