Answer to Question #100019 in Calculus for Santos

Question

Integration by parts

1. Integral of 5x sin4x dx

2. Integral of x lnx dx

Answer 1

Calculus

We need to find the Integrations using by parts

Solution:

$\int f(x) \space g(x) dx = f(x) \int g(x) dx -\int f' (x) (\int g(x) dx ) dx$

1)

$\int 5x \space sin \space 4x \space dx = 5 \int x \space sin \space 4x \space dx$

Here, f(x) = x and g(x) = sin 4x

Then $f ' (x) = 1$

$$\int 5x \space sin \space 4x \space dx =5 \Bigg( x \int sin 4x - \int 1. (\int sin 4x dx ) dx \Bigg)$$

$$= 5 \Bigg( x (\frac {-cos 4x }{4}) - \int (\frac {-cos 4x }{4}) dx \Bigg) + c$$

$$= 5 \Bigg( \frac {- x cos 4x}{4 } + \frac {sin 4x}{16} \Bigg) + c$$

2)

$\int x \space ln x \space dx = ln x \int x \space dx - \int d( ln x)( \int x dx ) dx$

$$= ln x \times \frac {x^2 }{2} - \int \frac {1}{x} \times \frac {x^2}{2} dx + c$$

$$= \frac {x^2}{2} \times ln x - \frac {1}{2} \int x dx + c$$

$$= \frac {x^2}{2} \times ln x - \frac {1}{2} \times \frac {x^2} {2} + c$$

$= = \frac {x^2}{2} \times ln x - \frac {x^2} {4} + c$