Question

Find the integrals (primitives):
$\rm \displaystyle\int \frac{\sin 4x}{\sin x}\ dx$

Answer 1

here is the answer ✌️✌️✍️✍️✍️

Answer 2

Answer:

$\rm \displaystyle\int \frac{\sin 4x}{\sin x}\ dx= 4sinx - \dfrac {8sin^3}3 + C$

Step-by-step explanation:

By the formula of sin2x = 2sinxcosx

$\rm \displaystyle\int \frac{\sin 4x}{\sin x}\ dx = \rm \displaystyle\int \frac{2\sin 2x cos2x}{\sin x}\ dx=\rm \displaystyle\int \frac{2.2\sin x \ cosx \ cos2x}{\sin x}\ dx$

By the formula of cos2x

$cos2x = 1 - 2sin^2x$

$\int 4\ cosx\ cos2x\ dx = 4\int cosx (1-2sin^2x)dx$

let u= sinx

$\dfrac{du}{dx} = cosx \\du = cosx\ dx$

Now,

$4\int(1-2u^2)du = 4\int du - 8\int u^2du\\\\ = 4u - \dfrac {8u^3}3 + C .....(1)$

Now put the value of u in equation (1)

$\rm \displaystyle\int \frac{\sin 4x}{\sin x}\ dx= 4sinx - \dfrac {8sin^3}3 + C$