Question

cos√x
--------- . dx
√x ​

Answer 1

$\sf \longrightarrow \displaystyle \int \rm \dfrac{cos \: \sqrt{x} }{ \sqrt{x} } dx \\ \\ \\ \rm let \: \sqrt{x} \: = t \\ \\ \rm \dfrac{1}{2} \: \frac{1}{\sqrt{x}} \: dx= dt\\ \\ \boxed{\rm \frac{1}{\sqrt{x}} \: dx=2 dt} \\ \\ \\ \sf \longrightarrow \displaystyle 2 \int \rm {cos \: t } \: dt\\ \\ \\ \sf \longrightarrow 2 \rm \: {sin\: t } + c \\ \\ \rm now \: put \: \bold{ t \: = \sqrt{x} }\\ \\ \\ \sf \longrightarrow 2 \rm \: {sin\: \sqrt{x}} + c$

Additional Information :-

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