# Thread: Integration by substitution and part

1. ## Integration by substitution and part

okay i got two question which i have real problems with. Any help would be great cheers guys

a)intergrate this using the substitution u = x^2 + 1

I = ∫2x(x^2 + 1)^3 dx

b) intergrate this using parts

J = ∫3xsinx dx

any help would be brilliant cheers guys

okay i got two question which i have real problems with. Any help would be great cheers guys

a)intergrate this using the substitution u = x^2 + 1

I = ∫2x(x^2 + 1)^3 dx
Well... $u = x^2 + 1 \implies du = 2xdx$

So
$I = \int (x^2 + 1)^3 \cdot 2x dx = \int u^3 du$

-Dan

b) intergrate this using parts

J = ∫3xsinx dx

any help would be brilliant cheers guys
$I = 3\int x sin(x) dx$

Let
$u = x \implies du = dx$
$dv = sin(x) dx \implies v = -cos(x)$

So
$I = 3\int x sin(x) dx = 3 \left ( -xcos(x) + \int cos(x) dx \right )$

-Dan

b) intergrate this using parts

J = ∫3xsinx dx
Here's another solution (I'm not gonna take the constant):


\begin{aligned}
\int {x\sin x\,dx} &= \int {\left( {\cos x - \cos x + x\sin x} \right)\,dx}\\
&= \int {\left[ {\cos x - \left( {\cos x - x\sin x} \right)} \right]\,dx}\\
&= \int {\left( {\sin x - x\cos x} \right)'\,dx}\\
&= \sin x - x\cos x + k
\end{aligned}