Who's up to the challenge? 5

Calculus Level 4

$\large \int_0^1 x^2 \cos x \sin x \, dx$

If the integral above is in the form of

$\dfrac {-A + B\sin C - D\cos C}E ,$

where $A,B,C,D$ and $E$ are positive integers, find the minimum value of $A+B+C+D+E$ .

Clarification : Angles are measured in radians.

1 solution

Hamza A
Feb 15, 2016

apply IBP

we then have

$u=x^2,dv=\cos{x}\sin{x}$

now we have

$\displaystyle\int { { x }^{ 2 }\cos { x } \sin { x } dx } =\frac{x^2sin^{2}{x}}{2}-\displaystyle\int{ x\sin^{2}{x}}dx$

applying IBP again

$u=x,dv=\sin^{2}{x}$ $\displaystyle\int{ x\sin^{2}{x}}dx=\frac{x}{2}(x-(\frac{\sin{x}}{2})-\frac{1}{2}\displaystyle\int{x-\frac{\sin{x}}{2}dx}$

$\frac{x}{2}(x-(\frac{\sin{x}}{2})-\frac{1}{2}(\frac{x^2}{2}+\frac{\cos{2x}}{4})$

simplifying we get

$\frac{x^2\sin^{2}{x}}{2}+\frac{1}{2}(\frac{x^2}{2}+\frac{\cos{2x}}{4})-\frac{x}{2}(x-\frac{1}{2}\sin{2x})$

evaluating from 0 to 1

we get

$\frac{-1+2\sin {2}-\cos{2}}{8}$

so $A+B+C+D+E=14$

Nice solution, my method was just a little different,

$x^2 \sin (x) \cos (x) =\frac{1}{2} x^2 \sin (2x)$

I made this substitution in the beginning.

Akshay Yadav - 5 years, 4 months ago

that would be the smarter way to do it :)

i'm bad at trig so i didn't think of that

lol

Hamza A - 5 years, 3 months ago