Question

Integral sin²x.sinx​

Answer 1

Answer:

sin³x

Step-by-step explanation:

sin²x * sinx = sin³x

Answer 2

Solution :

$:\implies \displaystyle \sf{y = \int sin^{2}(x)\cdot sin(x)dx}$

$\sf{Now,\:by\:using\:substituting\:the\:value\:of\:sin^{2}x,\:in\:the\:above\:equation, \:we\:get :-} \\ \\ \sf{sin^{2}(x) = 1 - cos^{2}(x)}$

$:\implies \displaystyle \sf{y = \int (1 - cos^{2}(x))\cdot sin(x)dx}$

$\textsf{Now, by substituting u = cos(x) and du = -sin(x)dx}$

$:\implies\displaystyle \sf{y = \int -(1 - u)du}$

$:\implies\displaystyle \sf{y = \int (u - 1)du}$

$:\implies\displaystyle \sf{y = \int udu - \int du}$

$\textsf{Now, by using the power rule of integration, we get :} \\ \\ \underline{\sf{Power\:rule\:of\: differentiation :-}} \\ \\ \displaystyle \sf \int x^{n}dx = \dfrac{x^{(n + 1)}}{n + 1} + c$

$:\implies\displaystyle \sf{y = \dfrac{u^{(2 + 1)}}{2 + 1} - \int du}$

$:\implies\displaystyle \sf{y = \dfrac{u^{3}}{3} - u + c}$

$:\implies\textsf{Now, by substituting the value of u in the above equation, we get :-}$

$:\implies\displaystyle \sf{y = \dfrac{cos^{3}(x)}{3} - cos(x) + c}$

$\displaystyle \boxed{\therefore \sf{\displaystyle \sf{y = \int sin^{2}(x)\cdot sin(x)dx} = \dfrac{cos^{3}(x)}{3} - cos(x) + c}}$