Question

Evaluate intergal 1/sin(x) +root3cos(x). DX

Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{sinx + \sqrt{3} cosx}$

On divide and multiply the denominator by 2, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{\dfrac{1}{2}sinx + \dfrac{ \sqrt{3} }{2}cosx}$

can be further rewritten as

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{sin\dfrac{\pi}{6}sinx + cos\dfrac{\pi}{6}cosx}$

We know,

$\boxed{\tt{ cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y) \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{cos\bigg(x - \dfrac{\pi}{6}\bigg) }$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm sec\bigg(x - \dfrac{\pi}{6}\bigg) \: dx$

We know,

$\boxed{\tt{ \displaystyle\int\rm secx \: = \: log |secx + tanx| + c \: }}$

So, using this result, we get

$\rm \:  =  \: \dfrac{1}{2}log\bigg |sec\bigg(x - \dfrac{\pi}{6}\bigg) + tan\bigg(x - \dfrac{\pi}{6}\bigg) \bigg| + c$

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LEARN MORE

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

Answer 2

Answer:

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{dx}{sinx + \sqrt{3} cosx}$

On divide and multiply the denominator by 2, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{\dfrac{1}{2}sinx + \dfrac{ \sqrt{3} }{2}cosx}$

can be further rewritten as

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{sin\dfrac{\pi}{6}sinx + cos\dfrac{\pi}{6}cosx}$

We know,

$\boxed{\tt{ cosx \: cosy \: + \: sinx \: siny \: = \: cos(x - y) \: }}$

So, using this identity, we get

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm \frac{dx}{cos\bigg(x - \dfrac{\pi}{6}\bigg) }$

$\rm \:  =  \: \dfrac{1}{2}\displaystyle\int\rm sec\bigg(x - \dfrac{\pi}{6}\bigg) \: dx$

We know,

$\boxed{\tt{ \displaystyle\int\rm secx \: = \: log |secx + tanx| + c \: }}$

So, using this result, we get

$\rm \:  =  \: \dfrac{1}{2}log\bigg |sec\bigg(x - \dfrac{\pi}{6}\bigg) + tan\bigg(x - \dfrac{\pi}{6}\bigg) \bigg| + c$

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LEARN MORE

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