Question

integrate : sin²x dx​

Answer 1

Answer:

$\boxed{\orange{\displaystyle\sf \int sin^2x = \dfrac{x}{2} - \dfrac{sin(2x)}{4} + C }}$

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 ,

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

Let I be the integral of sin²x , then ,

$\displaystyle\sf\dashrightarrow I = \int sin^2x .dx \\\\\displaystyle\sf\dashrightarrow I =\int \dfrac{ 1 - cos(2x)}{2} dx$

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

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

• 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 ,

$\displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\bigg( x + C - \int \dfrac{cos (u)}{2} du \bigg) \\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\bigg( x + C -\dfrac{1}{2}( \int cos(u) .du \bigg) \\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}{2}\bigg( x + C -\dfrac{1}{2} ( sin(u) + C ) \bigg) \\\\\displaystyle\sf\dashrightarrow I =\dfrac{1}{2} \bigg( x -\dfrac{ sin(2x)}{2} + C \bigg) \\\\\dashrightarrow\boxed{\pink{\displaystyle\sf I =\dfrac{x}{2} - \dfrac{sin(2x)}{4} + C }}$

Answer 2

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int \sf \: \dfrac{sinx}{sin(x - a)}$

• ☆ On adding and Subtracting 'a' in angle of numerator,

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin(x - a + a)}{sin(x - a)}$

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin \bigg((x - a )+ a \bigg)}{sin(x - a)}$

$\rm \:= \: \:\:\displaystyle\int \sf \: \dfrac{sin(x - a)cosa + sina \: cos(x - a)}{sin(x - a)}$

$\red{\bigg \{ \because \: sin(x + y) = sinxcosy + sinycosx\bigg \}}$

$\rm \:= \: \:\displaystyle\int \sf \: cosa \: dx \: + \: \displaystyle\int \sf \: sina \: cot(x - a) \: dx=∫cosadx+∫sinacot(x−a)dx$

$\red{\bigg \{ \because \:\dfrac{cosx}{sinx} = cotx \bigg \}}$

$\rm \:= \: \:(cosa) \: x +sina \: logsin(x - a) + c=(cosa)x+sinalogsin(x−a)+c$

$\begin{gathered}\red{\bigg \{ \because \: \displaystyle\int \sf \: kdx \: = \: x \: + \: c\bigg \}} \\ \red{\bigg \{ \because \: \displaystyle\int \sf \: cotx = log(sinx) + c\bigg \}}\end{gathered}$

Additional Information :-

$\green{\boxed{\bf{\displaystyle\int \sf \: sinx = - cosx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cosx = sinx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: tanx = log \: secx + c = - log \: cosx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cotx = log \: sinx + c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: secxtanx = secx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: cosecxcotx = - cosecx+ c}}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: {sec}^{2}x = tanx+ c }}}$

$\green{\boxed{\bf{\displaystyle\int \sf \: {cosec}^{2}x = - cotx+ c }}}$