Question

$\boxed{\mathtt \red{\int \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi}}$
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Answer 1

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

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{(3 \sin \phi-2) \cos \phi}{5-\cos ^{2} \phi-4 \sin \phi} d \phi$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{(3 \sin \phi-2) \cos \phi}{5-(1 - \sin ^{2} \phi)-4 \sin \phi} d \phi$

$\rm \:  =  \: \displaystyle\int\rm \frac{(3 \sin \phi-2) \cos \phi}{5-1 + \sin ^{2} \phi-4 \sin \phi} d \phi$

$\rm \:  =  \: \displaystyle\int\rm \frac{(3 \sin \phi-2) \cos \phi}{4 + \sin ^{2} \phi-4 \sin \phi} d \phi$

can be re-arranged as

$\rm \:  =  \: \displaystyle\int\rm \frac{(3 \sin \phi-2) \cos \phi}{\sin ^{2} \phi-4 \sin \phi + 4} d \phi$

To evaluate this integral, we use method of Substitution

So, Substitute

$\red{\rm :\longmapsto\:sin \phi \: = \: y}$

$\red{\rm :\longmapsto\:cos\phi \:d \phi = \: dy}$

So, on substituting the values, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{(3y - 2)}{ {y}^{2} - 4y + 4 } \: dy$

$\rm \:  =  \: \displaystyle\int\rm \frac{(3y - 2)}{ {(y - 2)}^{2}} \: dy$

can be rewritten as

$\rm \:  =  \: \displaystyle\int\rm \frac{3(y - 2 + 2)- 2}{ {(y - 2)}^{2}} \: dy$

$\rm \:  =  \: \displaystyle\int\rm \frac{3(y - 2) + 6- 2}{ {(y - 2)}^{2}} \: dy$

$\rm \:  =  \: \displaystyle\int\rm \frac{3(y - 2) + 4}{ {(y - 2)}^{2}} \: dy$

$\rm \:  =  \: 3\displaystyle\int\rm \frac{1}{y - 2}dy + 4\displaystyle\int\rm \frac{1}{ {(y - 2)}^{2} }$

We know

$\boxed{\tt{ \displaystyle\int\rm \frac{1}{x}dx = log |x| + c}}$

and

$\boxed{\tt{ \displaystyle\int\rm {x}^{n} dx= \frac{ {x}^{n + 1} }{n + 1} + c}}$

So, using this, we get

$\rm \:  =  \: 3 \: log |y - 2| + 4\displaystyle\int\rm {(y - 2)}^{ - 2} \: dy$

$\rm \:  =  \: 3 \: log |y - 2| + 4\dfrac{ {(y - 2)}^{ - 2 + 1} }{ - 2 + 1} + c$

$\rm \:  =  \: 3 \: log |y - 2| + 4\dfrac{ {(y - 2)}^{ - 1} }{ - 1} + c$

$\rm \:  =  \: 3 \: log |y - 2| - \dfrac{4}{y - 2} + c$

$\rm \:  =  \: 3 \: log |sin \phi- 2| - \dfrac{4}{sin \phi - 2} + c$

$\rm \:  =  \: 3 \: log |sin \phi- 2| - 4(cosec\phi - 2)+ c$

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MORE TO KNOW

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