Question

sin wt× cos wt= ? in integration​

Answer 1

AnswEr :

Given Expression,

$\displaystyle \sf \: l = \int sin(wt)cos(wt).dt$

Multiplying and Dividing by 2,

$\implies \displaystyle \sf \: l = \dfrac{1}{2} \int 2sin(wt)cos(wt).dt$

Since, sin2∅ = 2sin∅cos∅

$\implies \displaystyle \sf \: l = \dfrac{1}{2} \int sin(2wt).dt$

Let u = 2wt.

Differentiating w.r.t t on both sides,

$\longrightarrow \sf \dfrac{du}{dt} = 2w \\ \\ \longrightarrow \sf \dfrac{du}{2w} = dt$

Now,

$\implies \displaystyle \sf \: l = \dfrac{1}{2(2w)} \int sin(u).du \\ \\ \implies \displaystyle \sf \: l = - \dfrac{cos(t)}{2(2w)} + c \\ \\ \implies \boxed{ \boxed{ \sf \: l = -\dfrac{cos(wt)}{4w} + c}}$