Question

integrate sin (x - a)÷sin (x + a)dx
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Answer 1

Answer:

Step-by-step explanation:

Answer 2

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Integrate sin (x - a)÷sin (x + a)dx

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$∫sin 2 x= 2x​ − 4sin(2x)​ +C$

Step-by-step explanation:

We need to find the integral of sin²x . We know that cos 2x = 1 - 2sin²x . So , sin²x can be written as ,

$\sf\red{\dashrightarrow }sin^2x =\dfrac{ 1 - cos(2x)}{2}⇢sin 2 x= 21−cos(2x)$

Let I be the integral of sin²x , then ,

$\begin{gathered}\displaystyle\sf\dashrightarrow I = \int sin^2x .dx \\\\\displaystyle\sf\dashrightarrow I =\int \dfrac{ 1 - cos(2x)}{2} dx \end{gathered} ⇢I=∫sin 2 x.dx⇢I=∫ 21−cos(2x)​ dx$

Now here 1/2 is a constant with respect to x we can move it out of the integral.

$\begin{gathered}\displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\int 1 - cos(2x) .dx \\\\\displaystyle\sf\dashrightarrow I = \dfrac{1}{2}\bigg( \int 1.dx +\int cos (2x) .dx\bigg) \\\\\displaystyle\sf\dashrightarrow I = \dfrac{1}{2}\bigg( x + C + \int cos (2x) \bigg).dx \end{gathered} ⇢I= 21​ ∫1−cos(2x).dx⇢I= 21​ (∫1.dx+∫cos(2x).dx)⇢I= 21​ (x+C+∫cos(2x)).dx$

Now here , let us take u = 2x . Therefore on differenciating both sides wrt x , we will get du/dx = 2 . So that , dx = du/2 . On substituting this ,

$\begin{gathered} \displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\bigg( x + C - \int \dfrac{cos (u)}{2} du \bigg) \\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}$${2}\bigg( x + C -\dfrac{1}{2}( \int cos(u) .du \bigg)$

$\\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\bigg( x + C -\dfrac{1}{2} ( sin(u) + C ) \bigg) \\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}{2} \bigg( x -\dfrac{ sin(2x)}{2} + C \bigg)$$\\\\\dashrightarrow\boxed{\pink{\displaystyle\sf I =\dfrac{x}{2} - \dfrac{sin(2x)}{4} + C }}\end{gathered} ⇢I$$= 21​ (x+C−∫ 2cos(u)​ du)⇢I= 21\\​ (x+C− 21​ (∫cos(u).du)⇢I= 21​ (x+C− 21​ (sin(u)+C))⇢I= 21​ (x− 2sin(2x)​ +C)⇢ I= 2x​ − 4sin(2x)​ +C$

Thanks!