Question

integrate 2cosx-3sinx/6cosx+4sinx​

Answer 1

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

The given integral is

$\displaystyle\int\rm \frac{2cosx - 3sinx}{6cosx + 4sinx} \: dx$

can be rewritten as

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

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm \: 3cosx + 2sinx = y$

$\rm \: ( - 3sinx + 2cosx) \: dx \: = \: dy$

$\rm \: (2cosx - 3sinx) \: dx \: = \: dy$

So, on substituting these values in above integral, we get

$\rm \: = \: \dfrac{1}{2}\displaystyle\int\rm \frac{dy}{y}$

$\rm \: = \: \dfrac{1}{2}log |y| + c$

$\rm \: = \: \dfrac{1}{2}log |3cosx + 2sinx| + c$

Hence,

$\boxed{\tt{ \rm \: \displaystyle\int\rm \frac{2cosx - 3sinx}{6cosx + 4sinx}dx = \: \dfrac{1}{2}log |3cosx + 2sinx| + c}}$

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FORMULA USED

$\boxed{\tt{ \dfrac{d}{dx}sinx = cosx}}$

$\boxed{\tt{ \dfrac{d}{dx}cosx = - \: sinx}}$

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

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ADDITIONAL INFORMATION

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

Answer 2

Question:-

Integrate the Function:

$\sf \large \frac{2cos \: x - 3sin \: x}{6cos \: x + 4sin \: x}$

Given:-

$\sf \large \frac{2cos \: x - 3sin \: x}{6cos \: x + 4sin \: x}$

Solution:-

Let, 3cos x + 2sin x = t.

∴ ( – 3sin x + 2cos x ) dx = dt.

$\sf \large \int \frac{2cos \: x - 3sin \: x}{6cos \: x + 4sin \: x} dx = \int \frac{dt}{2t}$

$\sf \large \frac{1}{2} \int \frac{1}{t} dt$

$\sf \large = \frac{1}{2} log |t| + c$

$\sf \large = \frac{1}{2} log |2sin \: x + 3cos \: x| + c.$

Answer:-

${ \boxed {{\sf \large \red{\frac{2cos \: x - 3sin \: x}{6cos \: x + 4sin \: x} \sf \large = \frac{1}{2} log |2sin \: x + 3cos \: x| + c.}}}}$

Hope you have satisfied. ⚘