Question

integrate : sin²x dx​

Answer 1

Answer:

You cannot directly integrate sin^2(x). Use trigonometric identities and calculus substitution rules to solve the problem. Use the half angle formula, sin^2(x) = 1/2*(1 - cos(2x)) and substitute into the integral so it becomes 1/2 times the integral of (1 - cos(2x)) dx.

Answer 2

Answer:

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

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 ,

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

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 ,

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