Question

Solve the following

$\int \: \frac{sin3x}{sin5x \: sin2x} \: dx$
​

Answer 1

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

Given integral is

$\rm \: \displaystyle\int\rm \frac{sin3x}{sin5x \: sin2x} \: dx$

can be rewritten as

$\rm \: = \: \displaystyle\int\rm \frac{sin(5x - 2x)}{sin5x \: sin2x} \: dx$

We know,

$\color{green}\boxed{ \rm{ \:sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }}$

So, using this result, we get

$\rm \: = \: \displaystyle\int\rm \frac{sin5xcos2x - sin2xcos5x}{sin5x \: sin2x} \: dx$

$\rm \: = \: \displaystyle\int\rm \frac{sin5xcos2x}{sin5x \: sin2x} \: dx - \displaystyle\int\rm \frac{cos5xsin2x}{sin5xsin2x} \: dx$

$\rm \: = \: \displaystyle\int\rm cot2x \: dx \: - \: \displaystyle\int\rm cot5x \: dx$

We know,

$\color{green}\boxed{ \rm{ \:\displaystyle\int\rm cotx \: dx \: = \: log |sinx| + c \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{log |sin2x| }{2} - \dfrac{log |sin5x| }{5} + c$

Hence,

$\color{green}\boxed{ \rm{ \:\displaystyle\int\rm \frac{sin3x}{sin5x \: sin2x} dx = \: \dfrac{log |sin2x| }{2} - \dfrac{log |sin5x| }{5} + c }}$

$\rule{190pt}{2pt}$

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

Step-by-step explanation:

Given integral is

\begin{gathered}\rm \: \displaystyle\int\rm \frac{sin3x}{sin5x \: sin2x} \: dx \\ \end{gathered}

∫

sin5xsin2x

sin3x

dx

can be rewritten as

\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{sin(5x - 2x)}{sin5x \: sin2x} \: dx \\ \end{gathered}

=∫

sin5xsin2x

sin(5x−2x)

dx

We know,

\begin{gathered}\color{green}\boxed{ \rm{ \:sin(x - y) = sinx \: cosy \: - \: siny \: cosx \: }} \\ \end{gathered}

sin(x−y)=sinxcosy−sinycosx

So, using this result, we get

\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{sin5xcos2x - sin2xcos5x}{sin5x \: sin2x} \: dx \\ \end{gathered}

=∫

sin5xsin2x

sin5xcos2x−sin2xcos5x

dx

\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{sin5xcos2x}{sin5x \: sin2x} \: dx - \displaystyle\int\rm \frac{cos5xsin2x}{sin5xsin2x} \: dx \\ \end{gathered}

=∫

sin5xsin2x

sin5xcos2x

dx−∫

sin5xsin2x

cos5xsin2x

dx

\begin{gathered}\rm \: = \: \displaystyle\int\rm cot2x \: dx \: - \: \displaystyle\int\rm cot5x \: dx \\ \end{gathered}

=∫cot2xdx−∫cot5xdx

We know,

\begin{gathered}\color{green}\boxed{ \rm{ \:\displaystyle\int\rm cotx \: dx \: = \: log |sinx| + c \: }} \\ \end{gathered}

∫cotxdx=log∣sinx∣+c

So, using this result, we get

\begin{gathered}\rm \: = \: \dfrac{log |sin2x| }{2} - \dfrac{log |sin5x| }{5} + c \\ \end{gathered}

=

2

log∣sin2x∣

−

5

log∣sin5x∣

+c

Hence,

\begin{gathered}\color{green}\boxed{ \rm{ \:\displaystyle\int\rm \frac{sin3x}{sin5x \: sin2x} dx = \: \dfrac{log |sin2x| }{2} - \dfrac{log |sin5x| }{5} + c }}\\ \end{gathered}

∫

sin5xsin2x

sin3x

dx=

2

log∣sin2x∣

−

5

log∣sin5x∣

+c