Question

find x/4 integration0 (sinx -cosx +sec^2x )dx​

Answer 1

Answer:

$\mathrm{2 -\sqrt2 \ or\ \sqrt2(\sqrt2-1)}$

Step-by-step explanation:

Appropriate Question :

$\mathrm{\displaystyle \int \limits_0^{\frac{\pi}{4}} (sinx - cosx + sec^2x) \, dx}$

To find :

The value of the given integral

Solution :

$\longmapsto \mathrm{\displaystyle \int \limits_0^{\frac{\pi}{4} (sinx - cosx + sec^2x) \, dx}$

We know that,

$\boxed{\begin{array}{cc} \bullet \mathrm{\displaystyle \int \sin {x} \, dx = -\cos x + c} \\\bullet \mathrm{\displaystyle \int \cos {x} \, dx = \sin x + c} \\\bullet \mathrm{\displaystyle \int \sec^2 {x} \, dx = \tan x + c} \end{array}}$

So, applying these results, we get,

$\implies \mathrm{\displaystyle \int \limits_0^{\frac{\pi}{4}} sinx \, dx - \displaystyle \int \limits_0^{\frac{\pi}{4}} cosx \, dx + \displaystyle \int \limits_0^{\frac{\pi}{4}} sec^2x \, dx}$

$\implies \mathrm{[-cosx]^{\frac{\pi}{4}}_0 - [sinx]^{\frac{\pi}{4}}_0 + [tanx]^{\frac{\pi}{4}}_0}$

$\implies \mathrm{-\bigg(\dfrac{1}{\sqrt2} - 1\bigg) -\bigg(\dfrac{1}{\sqrt2} - 0\bigg) + (1 - 0)}$

$\implies \mathrm{1 -\sqrt2 + 1 }$

$\implies \mathrm{2 -\sqrt2 \ or\ \sqrt2(\sqrt2-1)}$

$\therefore \underline{\boxed{\mathbf{\displaystyle \int \limits_0^{\frac{\pi}{4}} (sin x - cos x + sec^2x) \, dx = 2-\sqrt2}}}$

❖ Extra information :

⟡ Some basic integrals

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\\\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx+c \\\\ \sf x^n \ (n \neq -1)& \sf \dfrac{x^{n+1}}{n+1} + c \\\\ \sf \dfrac{1}{x} & \sf logx+ c\\\\ \sf {e}^{x} & \sf {e}^{x}+c\\\\ \sf sinx & \sf - \: cosx+ c \\\\ \sf cosx & \sf \: sinx + c\\\\ \sf {sec}^{2} x & \sf tanx + c\\\\ \sf {cosec}^{2}x & \sf - cotx+ c \\\\ \sf secx \: tanx & \sf secx + c\\\\ \sf cosecx \: cotx& \sf -\: cosecx + c\end{array}} \\ \end{gathered}$

Hope it helps!!