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Calculus Made Easy

(4) Find $\int x^{2}\sin x\,dx$ .

${\begin{aligned}{\text{Write }}\;x^{2}&=u,\quad \quad \sin x\,dx=dv;\\{\text{then }}\;du&=2x\,dx,\quad \quad v=-\cos x,\end{aligned}}$

$\int x^{2}\sin x\,dx=-x^{2}\cos x+2\int x\cos x\,dx$ .

Now find $\int x\cos x\,dx$ , integrating by parts (as in Example 1 above):

$\int x\cos x\,dx=x\sin x+\cos x+C$ .

Hence

${\begin{aligned}\int x^{2}\sin x\,dx&=-x^{2}\cos x+2x\sin x+2\cos x+C'\\&=2\left[x\sin x+\cos x\left(1-{\frac {x^{2}}{2}}\right)\right]+C'.\end{aligned}}$

(5) Find $\int {\sqrt {1-x^{2}}}\,dx$ .

${\begin{aligned}{\text{Write}}\;\quad \quad \quad u&={\sqrt {1-x^{2}}},\quad dx=dv;\\{\text{then }}\;\quad \quad \quad du&=-{\frac {x\,dx}{\sqrt {1-x^{2}}}}\quad {\text{(see Chap. IX., p. 67)}}\end{aligned}}$

and $x=v$ ; so that

$\int {\sqrt {1-x^{2}}}\,dx=x{\sqrt {1-x^{2}}}+\int {\frac {x^{2}\,dx}{\sqrt {1-x^{2}}}}$ .

Here we may use a little dodge, for we can write

$\int {\sqrt {1-x^{2}}}\,dx=\int {\frac {(1-x^{2})\,dx}{\sqrt {1-x^{2}}}}=\int {\frac {dx}{\sqrt {1-x^{2}}}}-\int {\frac {x^{2}\,dx}{\sqrt {1-x^{2}}}}$ .

Adding these two last equations, we get rid of $\int {\dfrac {x^{2}\,dx}{\sqrt {1-x^{2}}}}$ , and we have

$2\int {\sqrt {1-x^{2}}}\,dx=x{\sqrt {1-x^{2}}}+\int {\frac {dx}{\sqrt {1-x^{2}}}}$ .

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