Question

Evaluate the integrals in Exercises 91–102.102. ∫(from -1/3 to 1/√3)(cos(arctan 3x))/(1+9x²) dx

Accepted Answer

Recognize that the integral involves the expression $$ \cos(\arctan(3x)) $$ and the denominator $$ 1 + 9x^2 . $$This suggests a trigonometric substitution related to the angle $$ 	heta = \arctan(3x) . $$Set $$ 	heta = \arctan(3x) $$, which implies $$ 3x = 	an(	heta) . $$From this, express $$ x  in $$terms of $$ 	heta $$: $$ x = \frac{	an(	heta)}{3} . $$Calculate the differential $$ dx  in $$terms of $$ d	heta . $$Since $$ x = \frac{	an(	heta)}{3} $$, then $$ dx = \frac{1}{3} \sec^2(	heta) d	heta . $$Rewrite the integral limits in terms of $$ 	heta  by $$substituting the original $$ x $$ limits into $$ 	heta = \arctan(3x) . $$For $$ x = -\frac{1}{3} $$, $$ 	heta = \arctan(-1) = -\frac{\pi}{4} . $$For $$ x = \frac{1}{\sqrt{3}} $$, $$ 	heta = \arctan\left(3 \cdot \frac{1}{\sqrt{3}}ight) = \arctan(\sqrt{3}) = \frac{\pi}{3} . $$Substitute all parts into the integral: replace $$ \cos(\arctan(3x)) $$ with $$ \cos(	heta) $$, replace $$ 1 + 9x^2 $$ with $$ 1 + 	an^2(	heta) = \sec^2(	heta) $$, and replace $$ dx $$ with $$ \frac{1}{3} \sec^2(	heta) d	heta . $$Simplify the integrand and integral accordingly before integrating with respect to $$ 	heta $$ from $$ -\frac{\pi}{4}  to$$ $$ \frac{\pi}{3} .$$

Evaluate the integrals in Exercises 91–102.
102. ∫(from -1/3 to 1/√3)(cos(arctan 3x))/(1+9x²) dx

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