Concept:\(\displaystyle \int \ x^n\ dx = \frac{x^{n+1}}{n+1}+C, n\neq -1\)Calculation:\(\displaystyle \int^{\frac{?}{4}}_0 \frac{dx}{(\sin x + \cos x)^2}\)We know that \(\displaystyle tan\ x =\frac{sin\ x}{cos\ x}\) ? \(\displaystyle sin\ x =\frac{tan\ x}{sec\ x}\) and \(\displaystyle cos\ x =\frac{1}{sec\ x}\)? \(\displaystyle I= \int^{\frac{?}{4}}_0 \frac{dx}{(\frac{tan\ x}{sec\ x} +\frac{1}{sec\ x})^2}\)? \(\displaystyle I= \int^{\frac{?}{4}}_0 \frac{{sec^2\ x}\ dx}{(\tan x + 1)^2}\)Let tan x = t ? sec2x dx = dtWhen x = 0 ? t = tan 0 ? t = 0When x = ?/4 ? t = tan ?/4 ? t = 1Let us now substitute the 't' value and dt value in the given expression for integration.? \(\displaystyle I= \int^{1}_0 \frac{dt}{(1+t)^2}\)Let (1 + t) = v ? dv = dt When t = 0 ? v = 1 + 0 ? v = 1When t = 1 ? v = 1 + 1 ? v = 2Let us now substitute the 'v' value and dv value in the given expression for integration.? \(\displaystyle I= \int^{2}_1 \frac{dv}{v^2} \)Using \(\displaystyle \int \ x^n\ dx = \frac{x^{n+1}}{n+1}+C, n\neq -1\),? \(\displaystyle I=\left[ \frac{-1}{v}\right ]_1^2 \)? \(\displaystyle I=\left[ \frac{-1}{2}- \frac{-1}{1}\right ] \) = 1/2? \(\displaystyle \int^{\frac{?}{4}}_0 \frac{dx}{(\sin x + \cos x)^2}=\frac{1}{2}\)