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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.
Start Solution

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.
Show Step 2

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 3

Doing 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 $$u$$!
Show Step 4

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.