Question

Integration of sin4x/sinx

Answer 1

hope it helps .

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Answer 2

Answer:

Integration of $\bold{\frac{\sin 4 x}{\sin x}=\frac{2}{3} \sin 3 x+3 \sin x-4 \sin x+c}$

Step-by-step explanation:

$\begin{array}{l}{\int \frac{\sin 4 x}{\sin x}=\int \frac{2 \sin 2 x \cos 2 x}{\sin x}} \\ {\because \sin 2 x=2 \sin x \cos x} \\ {=\int \frac{2 \times 2 \sin x \cos x \times \cos 2 x}{\sin x}} \\ {=4 \int \frac{\sin x \cos x \times \cos 2 x}{\sin x}}\end{array}$

Cancelling sinx, we get,

$=4 \int \cos x \times \cos 2 x$

Here cos2x is written as $\bold{2 \cos ^{2} x-1}$ according to the formula,  

$\begin{array}{l}{=4 \int \cos x\left(2 \cos ^{2} x-1\right)} \\ {=4 \int 2 \cos ^{3} x \times \cos x}\end{array}$

Again using the formula,

$=\frac{8}{4} \int(\cos 3 x+3 \cos x)-4 \int \cos x$

[Since,$\cos 3 x=4 \cos ^{3} x-3 \cos x \Rightarrow \cos ^{3} x=\frac{\cos 3 x+3 \cos x}{4} ]$

$=2\left(\frac{\sin 3 x}{3}+3 \sin x\right)-4 \sin x+c$

$=\frac{2}{3} \sin 3 x+3 \sin x-4 \sin x+c$

∴Integration of $\bold{\frac{\sin 4 x}{\sin x}=\frac{2}{3} \sin 3 x+3 \sin x-4 \sin x+c}$