(4) Find ∫ x 2 sin x d x {\displaystyle \int x^{2}\sin x\,dx} .
Write x 2 = u , sin x d x = d v ; then d u = 2 x d x , v = − cos x , {\displaystyle {\begin{aligned}{\text{Write }}\;x^{2}&=u,\quad \quad \sin x\,dx=dv;\\{\text{then }}\;du&=2x\,dx,\quad \quad v=-\cos x,\end{aligned}}}
∫ x 2 sin x d x = − x 2 cos x + 2 ∫ x cos x d x {\displaystyle \int x^{2}\sin x\,dx=-x^{2}\cos x+2\int x\cos x\,dx} .
Now find ∫ x cos x d x {\displaystyle \int x\cos x\,dx} , integrating by parts (as in Example 1 above):
∫ x cos x d x = x sin x + cos x + C {\displaystyle \int x\cos x\,dx=x\sin x+\cos x+C} .
Hence
∫ x 2 sin x d x = − x 2 cos x + 2 x sin x + 2 cos x + C ′ = 2 [ x sin x + cos x ( 1 − x 2 2 ) ] + C ′ . {\displaystyle {\begin{aligned}\int x^{2}\sin x\,dx&=-x^{2}\cos x+2x\sin x+2\cos x+C'\\&=2\left[x\sin x+\cos x\left(1-{\frac {x^{2}}{2}}\right)\right]+C'.\end{aligned}}}
(5) Find ∫ 1 − x 2 d x {\displaystyle \int {\sqrt {1-x^{2}}}\,dx} .
Write u = 1 − x 2 , d x = d v ; then d u = − x d x 1 − x 2 (see Chap. IX., p. 67) {\displaystyle {\begin{aligned}{\text{Write}}\;\quad \quad \quad u&={\sqrt {1-x^{2}}},\quad dx=dv;\\{\text{then }}\;\quad \quad \quad du&=-{\frac {x\,dx}{\sqrt {1-x^{2}}}}\quad {\text{(see Chap. IX., p. 67)}}\end{aligned}}}
and x = v {\displaystyle x=v} ; so that
∫ 1 − x 2 d x = x 1 − x 2 + ∫ x 2 d x 1 − x 2 {\displaystyle \int {\sqrt {1-x^{2}}}\,dx=x{\sqrt {1-x^{2}}}+\int {\frac {x^{2}\,dx}{\sqrt {1-x^{2}}}}} .
Here we may use a little dodge, for we can write
∫ 1 − x 2 d x = ∫ ( 1 − x 2 ) d x 1 − x 2 = ∫ d x 1 − x 2 − ∫ x 2 d x 1 − x 2 {\displaystyle \int {\sqrt {1-x^{2}}}\,dx=\int {\frac {(1-x^{2})\,dx}{\sqrt {1-x^{2}}}}=\int {\frac {dx}{\sqrt {1-x^{2}}}}-\int {\frac {x^{2}\,dx}{\sqrt {1-x^{2}}}}} .
Adding these two last equations, we get rid of ∫ x 2 d x 1 − x 2 {\displaystyle \int {\dfrac {x^{2}\,dx}{\sqrt {1-x^{2}}}}} , and we have
2 ∫ 1 − x 2 d x = x 1 − x 2 + ∫ d x 1 − x 2 {\displaystyle 2\int {\sqrt {1-x^{2}}}\,dx=x{\sqrt {1-x^{2}}}+\int {\frac {dx}{\sqrt {1-x^{2}}}}} .