Question

Integrate the following function with respect to x

$\sqrt{1 - sin2x} \: when \: x \: \in \: (0 \: to \: \frac{\pi}{4})$

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Answer 1

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

Given integral is

$\rm :\longmapsto\: \: \displaystyle\int\rm \sqrt{1 - sin2x} \: dx$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \sqrt{1 - 2sinx \: cosx} \: dx$

can be further rewritten as

$\rm \:  =  \: \displaystyle\int\rm \sqrt{ {sin}^{2}x + {cos}^{2}x - 2sinx \: cosx} \: dx$

$\rm \:  =  \: \displaystyle\int\rm \sqrt{ {(sinx - cosx)}^{2} } \: dx$

$\rm \:  =  \: \displaystyle\int\rm |sinx - cosx| \: dx$

Now, We know, By definition of Modulus function

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |x| = \begin{cases} &\sf{ - x \: when \: x < 0} \\ \\ &\sf{ \: \: x \: when \: x \geqslant 0} \end{cases}\end{gathered}\end{gathered}$

Now,

$\begin{gathered}\begin{gathered}\bf\: \rm :\longmapsto\: |sinx - cosx| = \begin{cases} &\sf{ - (sinx - cosx) \: when \: 0 < x < \dfrac{\pi}{4} } \\ \\ &\sf{ \: \: sinx - cosx \: when \: \dfrac{\pi}{4} \leqslant x < \dfrac{\pi}{2}} \end{cases}\end{gathered}\end{gathered}$

So, using this, we can rewrite the above integral as

$\rm \:  =  \: \displaystyle\int\rm - (sinx - cosx) \: dx$

$\rm \:  =  \: \displaystyle\int\rm (cosx - sinx) \: dx$

$\rm \:  =  \: sinx + cosx + c$

Hence,

$\rm\implies \:\boxed{\tt{ \rm \: \displaystyle\int\rm \sqrt{1 - sin2x}dx =  \: sinx + cosx + c}}$

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MORE TO KNOW

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