Question

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$\implies \: \red{ \int \:x \sin(\pi \: x)dx}$



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Answer 1

Answer:

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

Step-by-step-explanation:

We have to integrate the given function.

The given function is

$\displaystyle{\sf\:x\:\sin\:(\:\pi\:x\:)}$

Now, using integration by parts,

$\displaystyle{\boxed{\pink{\sf\:\int\:(\:uv'\:)\:=\:uv\:-\:\int\:(\:u'v\:)\:}}}$

• u is a function
• v' is another function
• u' is derivative of u
• v is integral of v'

Let,

$\displaystyle{\sf\:x\:=\:u}$

$\displaystyle{\sf\:\sin\:(\:\pi\:x\:)\:=\:v'}$

Now,

$\displaystyle{\sf\:u'\:=\:\dfrac{d}{dx}\:(\:u\:)}$

$\displaystyle{\implies\sf\:u'\:=\:\dfrac{d}{dx}\:(\:x\:)}$

$\displaystyle{\implies\:\boxed{\sf\:u'\:=\:1}}$

Now,

$\displaystyle{\sf\:v\:=\:\int\:v'}$

$\displaystyle{\implies\sf\:v\:=\:\int\:[\:\sin\:(\:\pi\:x\:)\:] \:dx}$

By substituting $\displaystyle{\sf\:\pi\:x\:=\:w}$,

Differentiating both sides w.r.t x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:w\:)\:=\:\dfrac{d}{dx}\:(\:\pi\:x\:)}$

$\displaystyle{\implies\sf\:\dfrac{dw}{dx}\:=\:\pi\:\dfrac{d}{dx}\:(\:x\:)}$

$\displaystyle{\implies\sf\:\dfrac{dw}{dx}\:=\:\pi\:\times\:1}$

$\displaystyle{\implies\sf\:\dfrac{dw}{dx}\:=\:\pi}$

$\displaystyle{\implies\sf\:dx\:=\:\dfrac{1}{\pi}\:dw}$

Now,

$\displaystyle{\sf\:v\:=\:\int\:\sin\:(\:w\:)\:\dfrac{1}{\pi}\:dw}$

$\displaystyle{\implies\sf\:v\:=\:\dfrac{1}{\pi}\:\int\:\sin\:(\:w\:)\:dw}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:\int\:\sin\:(\:x\:)\:dx\:=\:-\:\cos\:x\:+\:C\:}}}$

$\displaystyle{\implies\sf\:v\:=\:\dfrac{1}{\pi}\:(\:-\:\cos\:w\:)}$

Re-substituting $\displaystyle{\sf\:w\:=\:\pi\:x}$,

$\displaystyle{\implies\sf\:v\:=\:\dfrac{1}{\pi}\:[\:-\:\cos\:(\:\pi\:x\:)\:]}$

$\displaystyle{\implies\:\boxed{\sf\:v\:=\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}}}$

Now, using integration by parts,

$\displaystyle{\sf\:\int\:(\:uv'\:)\:=\:uv\:-\:\int\:(\:u'v\:)}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:x\:\left(\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}\:\right)\:-\:\int\:\left(\:1\:\times\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}\:\right)\:dx}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:x\:\cos\:(\:\pi\:x\:)}{\pi}\:-\:\int\:\left(\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}\:\right)\:dx}$

Now,

$\displaystyle{\sf\:\int\:\left(\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}\:\right)\:dx}$

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi}\:\int\:\cos\:(\:\pi\:x\:)\:dx}$

By substituting $\displaystyle{\sf\:\pi\:x\:=\:a}$,

Differentiating both sides w.r.t x, we get,

$\displaystyle{\sf\:\dfrac{d}{dx}\:(\:a\:)\:=\:\dfrac{d}{dx}\:(\:\pi\:x\:)}$

$\displaystyle{\implies\sf\:\dfrac{da}{dx}\:=\:\pi\:\dfrac{d}{dx}\:(\:x\:)}$

$\displaystyle{\implies\sf\:\dfrac{da}{dx}\:=\:\pi\:\times\:1}$

$\displaystyle{\implies\sf\:\dfrac{da}{dx}\:=\:\pi}$

$\displaystyle{\implies\sf\:dx\:=\:\dfrac{1}{\pi}\:da}$

Now,

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi}\:\int\:\cos\:(\:\pi\:x\:)\:dx}$

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi}\:\int\:\cos\:(\:a\:)\:\dfrac{1}{\pi}\:da}$

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi}\:\times\:\dfrac{1}{\pi}\:\int\:\cos\:(\:a\:)\:da}$

We know that,

$\displaystyle{\boxed{\green{\sf\:\int\:\cos\:(\:x\:)\:dx\:=\:\sin\:x\:+\:C\:}}}$

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi^2}\:\sin\:(\:a\:)}$

Re-substituting $\displaystyle{\sf\:a\:=\:\pi\:x}$,

$\displaystyle{\implies\sf\:\dfrac{-\:1}{\pi^2}\:\sin\:(\:\pi\:x\:)}$

Substituting this value in the original integral,

$\displaystyle{\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:x\:\cos\:(\:\pi\:x\:)}{\pi}\:-\:\int\:\left(\:\dfrac{-\:\cos\:(\:\pi\:x\:)}{\pi}\:\right)\:dx}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:x\:\cos\:(\:\pi\:x\:)}{\pi}\:-\:\left[\:\dfrac{-\:1}{\pi^2}\:\sin\:(\:\pi\:x\:)\:\right]}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:x\:\cos\:(\:\pi\:x\:)}{\pi}\:+\:\dfrac{\sin\:(\:\pi\:x\:)}{\pi^2}}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:x\:\cos\:(\:\pi\:x\:)}{\pi}\:\times\:\dfrac{\pi}{\pi}\:+\:\dfrac{\sin\:(\:\pi\:x\:)}{\pi^2}}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:\pi\:x\:\cos\:(\:\pi\:x\:)}{\pi^2}\:+\:\dfrac{\sin\:(\:\pi\:x\:)}{\pi^2}}$

$\displaystyle{\implies\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{-\:\pi\:x\:\cos\:(\:\pi\:x\:)\:+\:\sin\:(\:\pi\:x\:)}{\pi^2}}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:\int\:x\:\sin\:(\:\pi\:x\:)\:dx\:=\:\dfrac{\sin\:(\:\pi\:x\:)\:-\:\pi\:x\:\cos\:(\:\pi\:x\:)}{\pi^2}\:+\:C\:}}}}$