Question

Integrate it, please give the full solution.
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Answer 1

Consider the product rule of differentiation.

$(fg)'=f'g+fg'$

From this, we get,

$\displaystyle\int (f'g+fg')\ =\ fg\\ \\ \\ \Longrightarrow\ \int f'g\ +\ \int fg'\ =\ fg\ \ \ \ \ \longrightarrow\ \ \ \ \ (1)$

$\cline{1-}$

We have to integrate  $x\cdot\sin^3x.$

We know that,

$\sin(3x)=3\sin x-4\sin^3x$

This implies,

$\sin^3x=\dfrac{3\sin x-\sin(3x)}{4}$

So,

$\displaystyle\int x\cdot\sin^3x\ dx\ =\ \int\dfrac{x(3\sin x-\sin(3x))}{4}\ dx\\ \\ \\ =\dfrac{1}{4}\int x(3\sin x-\sin(3x))\ dx\\ \\ \\ =\dfrac{1}{4}\left[\int x\cdot3\sin x\ dx\ -\ \int x\cdot\sin(3x)\ dx\right]$

Now consider each integral in this equation.

Consider $\displaystyle\int x\cdot3\sin x\ dx.$

Using (1), let,

$f=x\ \ \ \Longrightarrow\ \ \ f'=1\\ \\ \\ g'=3\sin x\ \ \ \Longrightarrow\ \ \ g=-3\cos x$

So,

$\displaystyle\int f'g\ +\ \int fg'\ =\ fg\\ \\ \\ \Longrightarrow\ \int-3\cos x\ dx\ +\ \int x\cdot3\sin x\ dx\ =\ -3x\cdot\cos x\\ \\ \\ \Longrightarrow\ \int x\cdot3\sin x\ dx\ =\ -3x\cdot\cos x\ -\ \left(-3\int \cos x\ dx\right)\\ \\ \\ \Longrightarrow\ \int x\cdot3\sin x\ dx\ =\ 3\sin x - 3x\cdot\cos x\ =\ 3(\sin x-x\cdot\cos x)$

Consider  $\displaystyle\int x\cdot\sin(3x)\ dx.$

Using (1), let,

$\displaystyle f=x\ \ \ \Longrightarrow\ \ \ f'=1\\ \\ \\ g'=\sin(3x)\\ \\ \\ \Longrightarrow\ \ \ g=\int \sin(3x)\ dx\ =\ \dfrac{1}{3}\int \sin(3x)\ d(3x)\ =\ -\dfrac{1}{3}\cos(3x)$

So,

$\displaystyle\int f'g\ +\ \int fg'\ =\ fg\\ \\ \\ \Longrightarrow\ -\dfrac{1}{3}\int \cos(3x)\ dx\ +\ \int x\cdot\sin(3x)\ dx\ =\ -\dfrac{x}{3}\ \cos(3x)\\ \\ \\ \Longrightarrow\ \ \int x\cdot\sin(3x)\ dx\ =\ \dfrac{1}{9}\int\cos(3x)\ d(3x)\ -\ \dfrac{x}{3}\ \cos(3x)\\ \\ \\ \Longrightarrow\ \int x\cdot\sin(3x)\ dx\ =\ \dfrac{1}{9}\ \sin(3x)\ -\ \dfrac{3x}{9}\ \cos(3x)\\ \\ \\ \Longrightarrow\ \int x\cdot\sin(3x)\ dx\ =\ \dfrac{\sin(3x)-3x\cdot\cos(3x)}{9}$

Now,

$\displaystyle\dfrac{1}{4}\left[\int x\cdot3\sin x\ dx\ -\ \int x\cdot\sin(3x)\ dx\right]\\ \\ \\ \Longrightarrow\ \dfrac{1}{4}\left[3(\sin x-x\cdot\cos x)-\dfrac{\sin(3x)-3x\cdot\cos(3x)}{9}\right]\\ \\ \\ \Longrightarrow\ \dfrac{1}{4}\left[\dfrac{27\sin x-27x\cdot\cos x-\sin(3x)+3x\cdot\cos(3x)}{9}\right]\\ \\ \\ \Longrightarrow\ \dfrac{27\sin x-27x\cdot\cos x-\sin(3x)+3x\cdot\cos(3x)}{36}$

Hence the answer is,

$\boxed{\displaystyle\int x\cdot\sin^3x\ dx\ =\ \dfrac{27\sin x-27x\cdot\cos x-\sin(3x)+3x\cdot\cos(3x)}{36}+c}$

Answer 2

Step-by-step explanation:

We know that Int (u(x)*v(x)) dx = u(x)Int (v(x) - Int {u'(x)*Intv(x)dx}dx

Also we know that sin3x = 3sinx -4sin^3x.

Therefore sin^3x = (3sinx-sin3x)/4

We use the above results to integrate xsin^3x = x(3sinx-sin3x)/4.

u(x) = x and v(x) = (3sinx -sin3x)/4

u'x) = 1 and Int v(x) = Int {(3sinx - sin3x)/4 }dx = {-3cosx - (-cos3x/3)/4} = (-9cosx +cos3x)/12.

Therefore Int x sin^3x dx = Int { x* Int (sinx-sin3x)dx/4 - Int (x'*Int(sinx-sin3xdx) dx

=x(-9cosx+cos3x)/12 - Int (1*-9cosx+cos3xdx/12)

= x(-9cosx+cos3x)/12 +9sinx/12 - sin3x/36

Therefore Int xsin^3x = x(cos3x-9cosx)/12 + (3sinx)/4 - sin3x/36 .

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