Use the Table of Integrals to evaluate the integral.

Question

Quizplus · Accepted Answer

By using the Table of Integrals, we can see that the integral of $\frac{1}{\sqrt{1-x^2}}$ is $\arcsin{x}$ and the integral of $\cos^3{x}$ is $\frac{3}{4}\sin{x}-\frac{1}{4}\sin{3x}$. Therefore, using a trigonometric substitution of $x=\sin{u}$, we can rewrite the integral as $\int\frac{11}{\sqrt{1-\sin^2{u}}}\cos^3{u}\cos{u}\,du$. Simplifying this expression gives us $\int11\cos^4{u}\,du$. Using the identity $\cos^2{x}=\frac{1+\cos{2x}}{2}$, we can rewrite $\cos^4{u}$ as $\left(\cos^2{u}\right)^2=\left(\frac{1+\cos{2u}}{2}\right)^2=\frac{1+2\cos{2u}+\cos^2{2u}}{4}$. Substituting $v=2u$, we can rewrite the integral as $\frac{11}{4}\int(1+2\cos{v}+\cos^2{v})\,dv$. Using the identity $\cos^2{v}=\frac{1+\cos{2v}}{2}$ and integrating, we get $\frac{11}{4}\left(v+2\sin{v}+\frac{1}{2}v+\frac{1}{4}\sin{2v}\right)+C$. Substituting back $v=2u$, we get $\frac{11}{2}\left(2u+2\sin{2u}+\frac{1}{2}u+\frac{1}{4}\sin{4u}\right)+C$. Simplifying this expression gives us $\frac{11}{2}u+11\sin{2u}+\frac{11}{4}\sin{4u}+C$. Finally, substituting back $u=\arcsin{x}$, we get $\frac{11}{2}\arcsin{x}+11x\sqrt{1-x^2}+\frac{11}{4}(\sin{2\arcsin{x}})=\frac{11}{2}\arcsin{x}+11x\sqrt{1-x^2}+\frac{11}{4}(2x)=11x\sqrt{1-x^2}+\frac{11}{2}\arcsin{x}+\frac{11}{2}x$. Therefore, the answer is D.

Use the Table of Integrals to Evaluate the Integral