integral (xsin^2(x))dx
One helpful substitution is to note that sin^2(x) = (1-cos(2x))/2
That makes the problem
∫x(1-cos(2x))/2 dx = 1/2 ∫ (x - x cos2x) dx = 1/2 ∫x dx - 1/2 ∫x cos2x dx
The firrst part is trivial, and the second can be done using integration by parts. Let
u = x
dx
dv = cos2x dx
v = 1/2 sin2x
Then you have
∫ x cos 2x dx = 1/2 x sin2x - ∫1/2 sin2x dx
= 1/2 x sin2x + 1/4 cos2x
Now just put it all together.
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Remark:
sin ( x ) sin ( y ) = ( 1 / 2 ) [ cos ( y − x ) − cos ( y + x ) ]
sin² ( x ) = sin ( x ) ∙ sin ( x ) = ( 1 / 2 ) [ cos ( x − x ) − cos ( x + x ) ] =
( 1 / 2 ) [ cos ( 0 ) − cos ( 2x ) ] =
( 1 / 2 ) [ 1 − cos ( 2 x ) ]
So:
sin² ( x ) = ( 1 / 2 ) [ 1 − cos ( 2 x ) ]
To calculate the integral of \(x \sin^2(x) \, dx\), we can use integration by parts.
Integration by parts is a technique of integration that expresses the integral of a product of two functions as the product of one function with the integral of another function.
In this case, let's assign \(u = x\) and \(dv = \sin^2(x) \, dx\):
\(du = dx\) \quad (since the derivative of \(x\) with respect to \(x\) is 1)
\(v = \int \sin^2(x) \, dx\)
To find \(v\), we can rewrite \(\sin^2(x)\) in terms of trigonometric identities:
\(\sin^2(x) = \frac{1 - \cos(2x)}{2}\)
Now we can evaluate \(v\):
\(v = \int \sin^2(x) \, dx = \int \frac{1 - \cos(2x)}{2} \, dx\)
The integral of \(1 \, dx\) is \(x\), and the integral of \(-\cos(2x) \, dx\) can be found by substituting \(u = 2x\) and solving:
\(du = 2 \, dx \Rightarrow dx = \frac{du}{2}\)
Now the integral becomes:
\(v = \frac{1}{2} \int (1 - \cos(u)) \, du\)
\(= \frac{1}{2} \left(u - \sin(u)\right) + C\)
\(= \frac{1}{2} \left(2x - \sin(2x)\right) + C\)
Finally, applying the formula for integration by parts, we have:
\(\int x \sin^2(x) \, dx = u \cdot v - \int v \, du\)
\(= x \cdot \frac{1}{2} \left(2x - \sin(2x)\right) - \int \left(\frac{1}{2} \left(2x - \sin(2x)\right)\right) \, dx\)
\(= \frac{1}{2} \left(x^2 - x\sin(2x)\right) - \frac{1}{4} \left(x^2 - \frac{1}{2}\sin(2x) + C\right)\)
So, the solution to the integral is:
\(\int x \sin^2(x) \, dx = \frac{1}{2} \left(x^2 - x\sin(2x) - \frac{1}{2}\sin(2x)\right) + C\)
where \(C\) is the constant of integration.