Question

Find the integration of cosx with respect to x from 0 to π/2​

Answer 1

$\green{\large\underline{\sf{Solution-}}}$

$\rm :\longmapsto\:\displaystyle\int_{0}^{\dfrac{\pi}{2} } cosx \: dx$

We know,

$\boxed{\tt{ \displaystyle\int \: cosx \: dx \: = \: sinx \: + \: c \: }}$

So, using this, we get

$\rm \:  =  \: \bigg|sinx\bigg|_{0}^{\dfrac{\pi}{2} }$

$\rm \:  =  \: sin\dfrac{\pi}{2} - sin0$

$\rm \:  =  \: 1 - 0$

$\rm \:  =  \: 1$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int_{0}^{\dfrac{\pi}{2} } cosx \: dx = 1 \: }}$

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Additional Information :-

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