Question

integration of sin2x ​

Answer 1

Answer:

Questions

$\tt\int\sin2x \: dx$

Answer

Put 2x = t

$\implies \tt 2 = \dfrac{dt}{dx}$

$\implies\tt \: 2dx = dt$

$\implies \tt \: dx \: = \dfrac{1}{2} dt$

Put it, we get

$= \int \tt \sin t. \frac{1}{2} dt$

$= \tt \frac{1}{2} \int \tt \sin t \: dt$

$= \dfrac{1}{2}. (- \cos t) + c$

Now put the value of t we get ,

$\underline{\boxed{= \tt - \dfrac{1}{2} ( \cos2x) + c}}$

Some Integral Formulae

$\int \rm {x}^{n} dx = \dfrac{ {x}^{n + 1} }{n + 1} \: + c$

$\int \rm \cos x \: dx = \sin x + c$

$\int \rm \ \sin x \: dx = - \cos x + c$

$\int \rm { \sec }^{2} x \: dx = \tan x + c$

$\int \rm { \csc }^{2} x \: dx = - \cot x + c$

$\int \rm \sec x. \tan x \: dx= \sec x + c$

$\int \rm \csc x. \cot x \: dx= - \csc x+ c$

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