Question

find integration sin^2 x / 1+cos x DX​

Answer 1

Question :

$\displaystyle \int \sf\dfrac{sin^{2}(x)}{1 + cos(x)}dx$

Solution :

Let,

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

We know that, sin²x = (1 - cos²x), so by substituting it in the equation, we get :

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

Now, by using the identity of (a² - b²),i.e, (a - b)(a + b), we get :

$\sf I = \displaystyle \int \sf\dfrac{(1 - cos(x)(1 + cos(x))}{1 + cos(x)}dx$

By cancelling (1 + cosx), we get :

$\sf I = \displaystyle \int \sf 1 - cos(x)dx$

$\sf I = \displaystyle \int \sf 1dx - \displaystyle \int \sf cos(x)dx$

We know that, integral of dx is x and cosx is sinx, so by substituting them, we get :

$\sf I = x - sin(x) + C$

Hence,

$\displaystyle \int \sf\dfrac{sin^{2}(x)}{1 + cos(x)}dx = x - sin(x) + C$