How do you integrate #int_0^(pi/2) x^2*sin(2x) dx# using integration by parts?

Question

Emily Ammerman · Accepted Answer

#int_0^{pi/2} x^2*sin(2x) dx = frac{pi^2}{8} - 1/2#

#int f(x)*g'(x) dx = f(x)*g(x) - int f'(x)*g(x) dx#

To integrate by parts, you need to identify what is #f(x)# and #g'(x)#, i.e. which side to integrate and which side to differentiate.

Now, sine waves can be easily integrated or differentiated again, while polynomials can only be differentiated a finite number of times before becoming zero.

In this problem, you need to integrate by parts twice, because you need to differentiate #x^2# twice before it goes away. First, let #f(x)=x^2# and #g'(x)=sin(2x)#, i.e. differentiate #x^2# and integrate #sin(2x)#.

#int_0^{pi/2} x^2*sin(2x) dx = -1/2 int_0^{pi/2} x^2*frac{d}{dx}(cos(2x)) dx#

#= -1/2 ([x^2*cos(2x)]_0^{pi/2} - int_0^{pi/2} cos(2x)*frac{d}{dx}(x^2) dx)#

#= -1/2 [(pi/2)^2*(-1) - (0)^2*(1)] # #+ 1/2 int_0^{pi/2} cos(2x)*(2x) dx#

#= frac{pi^2}{8} + int_0^{pi/2} x*cos(2x) dx#

The second integration by parts is coming up now.

#frac{pi^2}{8} + int_0^{pi/2} x*cos(2x) dx = frac{pi^2}{8} + 1/2 int_0^{pi/2} x*frac{d}[dx}(sin(2x)) dx#

#= frac{pi^2}{8} + 1/2([x*sin(2x)]_0^{pi/2} - int_0^{pi/2} frac{d}[dx}(x)*sin(2x) dx)#

#= frac{pi^2}{8} + 1/2 [0-0] - 1/2 int_0^{pi/2} sin(2x) dx#

#= frac{pi^2}{8} - 1/2 int_0^{pi/2} sin(2x) dx#

#= frac{pi^2}{8} - 1/2 [-frac{cos(2x)}{2}]_0^{pi/2}#

#= frac{pi^2}{8} + 1/4 [cos(pi) - cos(0)]#

#= frac{pi^2}{8} - 1/2#

Samantha Adkison · Answer

undefined To integrate $ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx $ using integration by parts, follow these steps:

1. Choose $ u = x^2 $ and $ dv = \sin(2x) \, dx $.
2. Calculate $ du = 2x \, dx $ and $ v = -\frac{1}{2} \cos(2x) $.
3. Apply the integration by parts formula: $ \int u \, dv = uv - \int v \, du $.
4. Substitute the values of $ u $, $ v $, $ du $, and $ dv $ into the formula.
5. Evaluate the integrals and simplify the expression.

The integration by parts formula yields:

$$ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx = \left[ -\frac{1}{2}x^2 \cos(2x) \right]_{0}^{\frac{\pi}{2}} + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \cdot 2x \, dx $$

$$ = -\frac{1}{2}\left(\frac{\pi^2}{4}\right) \cos(\pi) - \left( -\frac{1}{2}(0)^2 \cos(0) \right) + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \cdot 2x \, dx $$

$$ = -\frac{\pi^2}{8} + \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2x) \cdot 2x \, dx $$

Now, integrate $ \int_{0}^{\frac{\pi}{2}} \cos(2x) \cdot 2x \, dx $ by parts again:

1. Choose $ u = 2x $ and $ dv = \cos(2x) \, dx $.
2. Calculate $ du = 2 \, dx $ and $ v = \frac{1}{2} \sin(2x) $.
3. Apply the integration by parts formula.
4. Evaluate the integrals and simplify the expression.

This yields:

$$ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx = -\frac{\pi^2}{8} + \left[ \frac{1}{2}x \sin(2x) \right]_{0}^{\frac{\pi}{2}} - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx $$

$$ = -\frac{\pi^2}{8} + \left( \frac{1}{2}\frac{\pi}{2} \sin(\pi) - \frac{1}{2}(0) \sin(0) \right) - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin(2x) \, dx $$

$$ = -\frac{\pi^2}{8} - \frac{\pi}{4} $$

Thus, $ \int_{0}^{\frac{\pi}{2}} x^2 \sin(2x) \, dx = -\frac{\pi^2}{8} - \frac{\pi}{4} $.