# How do you integrate int x^3 cos^2x dx  using integration by parts?

Jul 15, 2016

$= {x}^{4} / 8 + {x}^{3} / 4 \sin 2 x + \frac{3}{8} {x}^{2} \cos 2 x - \frac{3}{8} x \sin 2 x - \frac{3}{16} \cos 2 x + C$

#### Explanation:

that's very ugly and would need a load of IBP so i would start by making it even bigger!! using $\cos 2 A = 2 {\cos}^{2} A - 1$

$I = \int {x}^{3} {\cos}^{2} x \setminus \mathrm{dx}$

$= \int {x}^{3} \frac{\cos 2 x + 1}{2} \setminus \mathrm{dx}$

$\implies 2 I = \int {x}^{3} \cos 2 x + {x}^{3} \setminus \mathrm{dx}$

$\implies 2 I - {x}^{4} / 4 = \int {x}^{3} \cos 2 x \setminus \mathrm{dx} q \quad \triangle$ ......taking the easy bit to the LHS

so now we IBP the remaining bit over and over, reducing the ${x}^{3}$ term each time

$\int \setminus {x}^{3} \cos 2 x \setminus \mathrm{dx} = \int \setminus {x}^{3} \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \sin 2 x\right) \setminus \mathrm{dx}$

which by IBP....
$= {x}^{3} / 2 \sin 2 x - \int \setminus 3 {x}^{2} \frac{1}{2} \sin 2 x \setminus \mathrm{dx}$

$= {x}^{3} / 2 \sin 2 x - \frac{3}{2} \int \setminus {x}^{2} \sin 2 x \setminus \mathrm{dx}$

$= {x}^{3} / 2 \sin 2 x - \frac{3}{2} \int \setminus {x}^{2} \frac{d}{\mathrm{dx}} \left(- \frac{1}{2} \cos 2 x\right) \setminus \mathrm{dx}$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \int \setminus {x}^{2} \frac{d}{\mathrm{dx}} \left(\cos 2 x\right) \setminus \mathrm{dx}$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \left({x}^{2} \cos 2 x - \int \setminus 2 x \cos 2 x \setminus \mathrm{dx}\right)$

and on we go!!

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \left({x}^{2} \cos 2 x - \int \setminus 2 x \frac{d}{\mathrm{dx}} \left(\frac{1}{2} \sin 2 x\right) \setminus \mathrm{dx}\right)$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \left({x}^{2} \cos 2 x - \int \setminus x \frac{d}{\mathrm{dx}} \left(\sin 2 x\right) \setminus \mathrm{dx}\right)$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \left({x}^{2} \cos 2 x - \left(x \sin 2 x - \int \setminus \sin 2 x \setminus \mathrm{dx}\right)\right)$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} \left({x}^{2} \cos 2 x - \left(x \sin 2 x + \frac{1}{2} \cos 2 x\right)\right) + C$

$= {x}^{3} / 2 \sin 2 x + \frac{3}{4} {x}^{2} \cos 2 x - \frac{3}{4} x \sin 2 x - \frac{3}{8} \cos 2 x + C$

so back to $\triangle$

$\implies 2 I - {x}^{4} / 4 = {x}^{3} / 2 \sin 2 x + \frac{3}{4} {x}^{2} \cos 2 x - \frac{3}{4} x \sin 2 x - \frac{3}{8} \cos 2 x + C$

$\implies I = {x}^{4} / 8 + {x}^{3} / 4 \sin 2 x + \frac{3}{8} {x}^{2} \cos 2 x - \frac{3}{8} x \sin 2 x - \frac{3}{16} \cos 2 x + C$

once again , the key is IBP:

$\int \setminus u \left(x\right) \cdot \frac{d}{\mathrm{dx}} \left(v \left(x\right)\right) \setminus \mathrm{dx} = u \left(x\right) \cdot v \left(x\right) - \int \setminus \frac{d}{\mathrm{dx}} \left(u \left(x\right)\right) \cdot v \left(x\right) \setminus \mathrm{dx}$