Question

Please Help Urgently ​

Answer 1

Given Integrand,

$\displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} }sec \: x \sqrt{ \dfrac{1 - sin \: x}{1 + sin \: x} } dx$

Rationalising the denominator,

$\longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} }sec \: x \sqrt{ \dfrac{1 - sin \: x}{1 + sin \: x} \times\dfrac{1 - sin \: x}{1 - sin \: x} } dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} }sec \: x \sqrt{ \dfrac{(1 - sin \: x) {}^{2} }{1 - sin {}^{2} \: x}}dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} }sec \: x \times \dfrac{ \sqrt{(1 - sin \: x) {}^{2} } }{ \sqrt{ {cos}^{2} x}} dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} }sec \: x \times \dfrac{1 - sin \: x}{cos \: x} dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} } \dfrac{1 - sin \: x}{cos^2 \: x} dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = \int_{0}^{ \frac{\pi}{4} } sec^2x dx - \int_{0}^{ \frac{\pi}{4} }sec \: x tan \ x \: dx \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = (tan \ x - sec \: x) \bigg| _{0}^{ \frac{\pi}{4} } \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) =f( \dfrac{\pi}{4} ) - f(0) \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = tan(\frac{\pi}{4}) + sec( \frac{\pi}{4} ) - 0 + sec0 \\ \\ \longrightarrow \: \displaystyle \sf \: f(x) = 1 + \sqrt{2} - 0 + 1 \\ \\ \longrightarrow \: \displaystyle \boxed{ \boxed{\sf \: f(x) = 2 - \sqrt{2} }}$

Answer 2

Answer : 2-√2

[ kindly refer to the attached picture for stepwise solution ]

Additional Information :

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}$