Question

solve by using Rule of Integration by parts​

Answer 1

$\Large\bf\red{\underline{\underline{Question}}}:$

• $\displaystyle\bf\int x \sin 2x dx$

$\Large\sf\red{\underline{\underline{Solution}}}:$

Integrating by parts

So, now we know :

∫ UV dx =  U ∫Vdx - ∫ {(U)'∫Vdx}dx

Take x as first function (U)  and sin 2x as second function (V) .

$\displaystyle\bf\int \red{x \sin 2x\;dx} = x\int\sin 2x - \int \left\{ x'\int\sin 2x\;dx \right\} dx$

$= \displaystyle x \left(\blue{\dfrac{-\cos 2x }{2}}\right) - \int \left\{ (\green{1}) .\left(\blue{\dfrac{-\cos 2x }{2}}\right) \right\} dx$

$=\displaystyle -\dfrac{x}{2}\cos 2x +\dfrac{1}{2}\int \cos 2x\; dx$

$= -\dfrac{x}{2} \cos 2x + \dfrac{1}{2}. \dfrac{\sin 2x}{2} +\sf C$

$= \dfrac{\sin 2x}{4} -\dfrac{x}{2} \cos 2x+\sf C$

Henceforth,

• $\displaystyle\bf\int x \sin 2x\; dx = \dfrac{ \sin 2x}{4} -\dfrac{x\cos x}{2} +\sf C$
• Option B is correct answer.

$\Large\bf\red{\underline{\underline{Points\;to;know}}}:$

• ∫ sin x = -cos x + c
• ∫ cos x = sin x + c
• ∫ tan x = -log | cos x | + c = log |sec x | + c
• ∫ cot x = log | sin x | + c

• ∫ sec² x = tan x + c
• $\displaystyle\int \sf x^n = \dfrac{x^{n+1}}{n+1}$