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Section 5.3 : Substitution Rule for Indefinite Integrals
15. Evaluate \( \displaystyle \int{{\frac{6}{{7 + {y^2}}}\,dy}}\).
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Hint : Be careful with this substitution. The integrand should look somewhat familiar, so maybe we should try to put it into a more familiar form.
The integrand looks an awful lot like the derivative of the inverse tangent.
\[\frac{d}{{du}}\left( {{{\tan }^{ - 1}}\left( u \right)} \right) = \frac{1}{{1 + {u^2}}}\]So, let’s do a little rewrite to make the integrand look more like this.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \int{{\frac{6}{{7\left( {1 + \frac{1}{7}{y^2}} \right)}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + \frac{1}{7}{y^2}}}\,dy}}\]
Hint : One more little rewrite of the integrand should make this look almost exactly like the derivative the inverse tangent and the substitution should then be fairly obvious.
Let’s do one more rewrite of the integrand.
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\int{{\frac{1}{{1 + {{\left( {\frac{y}{{\sqrt 7 }}} \right)}^2}}}\,dy}}\]At this point we can see that the following substitution will work for us.
\[u = \frac{y}{{\sqrt 7 }}\hspace{0.5in} \to \hspace{0.5in}du = \frac{1}{{\sqrt 7 }}dy\hspace{0.5in} \to \hspace{0.5in}\,dy = \sqrt 7 du\] Show Step 3Doing the substitution and evaluating the integral gives,
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \frac{6}{{\sqrt 7 }}{\tan ^{ - 1}}\left( u \right) + c\]
Hint : Don’t forget that the original variable in the integrand was not \(u\)!
Finally, don’t forget to go back to the original variable!
\[\int{{\frac{6}{{7 + {y^2}}}\,dy}} = \frac{6}{7}\left( {\sqrt 7 } \right)\int{{\frac{1}{{1 + {u^2}}}\,du}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{6}{{\sqrt 7 }}{{\tan }^{ - 1}}\left( {\frac{y}{{\sqrt 7 }}} \right) + c}}\]Substitutions for inverse trig functions can be a little tricky to spot when you are first start doing them. Once you do enough of them however they start to become a little easier to spot.