Question

integration sin3x/sinx​

Answer 1

$\huge \: \mathfrak { \purple{solution}}$

TO SOLVE:

$\int \: \frac{sin3x}{sinx } dx$

As we know that

$\boxed{ \green{ \bold{sin3x = 3sinx - 4 {sin}^{3} x}}}$

$\leadsto \: \int \: \frac{3sinx - 4 {sin}^{3} x}{sinx} dx\\ \\ \leadsto \: \int \: \frac{3sinx}{sinx} dx - \int \frac{4 {sin}^{3}x }{sinx} dx \\ \\ \leadsto \int3dx - \int4 {sin}^{2} \: xdx$

Now use

$\boxed{ \red{ \bold{{sin}^{2} x = \frac{1 - cos2x}{2}} }}$

$\leadsto \: 3 \int \: dx - 4 \int \: \frac{1 - cos2x}{2} dx \\ \\ \leadsto \: 3 \int \: dx - \frac{4}{2} \int \: 1 - cos2x \: dx \\ \\ \leadsto \: 3 \int \: dx - 2( \int \: dx - \int \: cos2x \: dx) \\ \\ \leadsto \: 3x - 2( x - \frac{sin2x}{2} ) + c \\ \\ \leadsto \: 3x - 2x + sin2x + c$

Formula used :

$\boxed{ \blue{\bold{ \int \: dx = x} }}\\ \\ \boxed{ \bold{\blue{ \int \: cos \: a x = \: \frac{sin \: a x}{a} }}}$

Answer 2

Answer:

Final answer is $sin 2x + x + c$

Step-by-step explanation:

Integration is process to sum up some parts to finds the whole parts.

Given: The trigonometry equation is sin3x/sinx​

Find: Need to find the integration part of the equation sin3x/sinx​

Solution:

$\int\limits^a_b \frac{Sin3x}{Sinx} \, dx \\=Sin2x+\int\limits dx \\= sin 2x + x + c$

So,  the final answer is $sin 2x + x + c$

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