## University Calculus: Early Transcendentals (3rd Edition)

$$\int\sqrt{1+\sin^2(x-1)}\sin(x-1)\cos(x-1)dx=\frac{1}{3}\Big(1+\sin^2(x-1)\Big)^{3/2}+C$$
$$A=\int\sqrt{1+\sin^2(x-1)}\sin(x-1)\cos(x-1)dx$$ a) Set $u=x-1$, then $du=dx$ $$A=\int\sqrt{1+\sin^2u}\sin u\cos udu$$ Set $v=\sin u$, then $dv=\cos udu$ $$A=\int(\sqrt{1+v^2})vdv$$ Set $w=1+v^2$, then $dw=2vdv$, or $vdv=1/2dw$ $$A=\frac{1}{2}\int\sqrt wdw=\frac{1}{2}\int w^{1/2}dw$$ $$A=\frac{1}{2}\times\frac{w^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{w^{3/2}}{3}+C$$ We have $$w=1+v^2=1+\sin^2u=1+\sin^2(x-1)$$ Therefore, $$A=\frac{1}{3}\Big(1+\sin^2(x-1)\Big)^{3/2}+C$$ b) Set $u=\sin(x-1)$, then $du=\cos(x-1)(x-1)'dx=\cos(x-1)dx$ $$A=\int(\sqrt{1+u^2})udu$$ Set $v=1+u^2$, then $dv=2udu$, or $udu=1/2dv$ $$A=\frac{1}{2}\int\sqrt vdv=\frac{1}{2}\int v^{1/2}dv$$ $$A=\frac{1}{2}\times\frac{v^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{v^{3/2}}{3}+C$$ We have $$v=1+u^2=1+\sin^2(x-1)$$ Therefore, $$A=\frac{1}{3}\Big(1+\sin^2(x-1)\Big)^{3/2}+C$$ c) Set $u=1+\sin^2(x-1)$, then $$du=2\sin(x-1)\Big(\sin(x-1)\Big)'dx=2\sin(x-1)\cos(x-1)dx$$ $$\frac{1}{2}du=\sin(x-1)\cos(x-1)dx$$ $$A=\frac{1}{2}\int\sqrt udu=\frac{1}{2}\int u^{1/2}du$$ $$A=\frac{1}{2}\times\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{u^{3/2}}{3}+C$$ $$A=\frac{1}{3}\Big(1+\sin^2(x-1)\Big)^{3/2}+C$$