## Thomas' Calculus 13th Edition

$$1$$
\eqalign{ & \int_0^{\pi /2} {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx \cr & {\text{Use the substitution method}}{\text{. }} \cr & u = 1 + 3{\sin ^2}x,\,\,\,\,du = 6\sin x\cos xdx,\,\,\,\,\,dx = \frac{{du}}{{6\sin x\cos x}} \cr & {\text{Then}}{\text{,}} \cr & \int {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx = \int {\frac{{3\sin x\cos x}}{{\sqrt u }}} \left( {\frac{{du}}{{6\sin x\cos x}}} \right) \cr & = \frac{1}{2}\int {\frac{{du}}{{\sqrt u }}} \cr & = \frac{1}{2}\int {{u^{ - 1/2}}du} \cr & {\text{Integrate by the power rule}} \cr & = \frac{1}{2}\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right) + C \cr & = \sqrt u + C \cr & {\text{Write in terms of }}x;{\text{ replace }}u = 1 + 3{\sin ^2}x \cr & = \sqrt {1 + 3{{\sin }^2}x} + C \cr & \cr & \int_0^{\pi /2} {\frac{{3\sin x\cos x}}{{\sqrt {1 + 3{{\sin }^2}x} }}} dx = \left( {\sqrt {1 + 3{{\sin }^2}x} } \right)_0^{\pi /2} \cr & {\text{Evaluate}} \cr & = \sqrt {1 + 3{{\sin }^2}\left( {\frac{\pi }{2}} \right)} - \sqrt {1 + 3{{\sin }^2}\left( 0 \right)} \cr & = \sqrt {1 + 3} - \sqrt {1 + 0} \cr & = 2 - 1 \cr & = 1 \cr}