Question

integral sin cube x dx ka man gyat kijiye​

Answer 1

Answer:

$\large\bold\red{\frac{ \cos(3x) }{12} - \frac{3 \cos(x) }{4} + c}$

Step-by-step explanation:

We have to integrate,

$\displaystyle \int { \sin}^{3} x dx$

But,

We know that,

$\sin(3x) = 3 \sin(x) - 4 { \sin }^{3} x \\ \\ = > 4 { \sin }^{3} x = 3 \sin(x) - \sin(3x) \\ \\ = > { \sin }^{3} x = \frac{3 \sin(x) - \sin(3x) }{4}$

Substituting this value,

We get,

$= \displaystyle \int \frac{3 \sin(x) - \sin(3x) }{4} dx \\ \\ \\ = \displaystyle \int \frac{3 \sin(x) }{4} dx - \displaystyle \int \frac{ \sin(3x) }{4}dx \\ \\ \\ = \frac{3}{4} \displaystyle \int \sin(x) dx - \frac{1}{4} \sin(3x) dx \\ \\ \\ = \frac{3}{4} \times ( - \cos x) - \frac{1}{4} \times \frac{ (- \cos 3x) }{3} + c\\ \\ \\ = - \frac{3 \cos(x) }{4} + \frac{ \cos(3x) }{12} + c \\ \\ \\ = \large \bold{ \frac{ \cos(3x) }{12} - \frac{3 \cos(x) }{4} + c}$

where,

c is an integral constant.

Answer 2

We have to integrate,

$\int { \sin}^{3} dx$

But,

We know that,

$\begin{lgathered}\sin(3x) = 3 \sin(x) - 4 { \sin }^{3} x \\ \\ = 4 { \sin }^{3} x = 3 \sin(x) - \sin(3x) \\ \\ = { \sin }^{3} x = \frac{3 \sin(x) - \sin(3x) }{4}\end{lgathered}$

Substituting this value,

We get,

$\begin{lgathered}= \displaystyle \int \frac{3 \sin(x) - \sin(3x) }{4} dx \\ \\ \\ = \displaystyle \int \frac{3 \sin(x) }{4} dx - \displaystyle \int \frac{ \sin(3x) }{4}dx \\ \\ \\ = \frac{3}{4} \displaystyle \int \sin(x) dx - \frac{1}{4} \sin(3x) dx \\ \\ \\ = \frac{3}{4} \times ( - \cos x) - \frac{1}{4} \times \frac{ (- \cos 3x) }{3} + c\\ \\ \\ = - \frac{3 \cos(x) }{4} + \frac{ \cos(3x) }{12} + c \\ \\ \\ = \large \bold{ \frac{ \cos(3x) }{12} - \frac{3 \cos(x) }{4} + c}\end{lgathered}$