Question

Calculus- Integration using table of integrals and Trig subsitution

Answer 1

\[\int\limits y \sqrt{10+4y+4y^2}dy\]

Answer 2

I need help solving this please. I'm stumped

Answer 3

Completing the square, you have \(\sqrt{10+4y+4y^2}=\sqrt{4\left(y+\dfrac{1}{2}\right)^2+9}\). Substituting \(y+\dfrac{1}{2}=\dfrac{3}{2}\tan t\) should do the trick.
\[\int y\sqrt{4\left(y+\dfrac{1}{2}\right)^2+9}\,dy=\int\left(\frac{3}{2}\tan t-\frac{1}{2}\right)\sqrt{4\left(\frac{3}{2}\tan t\right)^2+9}\,\left(\frac{3}{2}\sec^2t\,dt\right)\]

Answer 4

i know it has \[\sin^{-1} \] in it

Answer 5

according to the chart

Answer 6

If I may suggest something: It's always better to derive the results usually presented on a chart than to just refer to a chart to solve all your problems. It helps you to understand where the integral formulas come from, and in my experience it's drastically reduced the number of mistakes I made when making trig substitutions.

Answer 7

The above integral can be reduced quite a bit:
\[\frac{9}{4}\int\left(3\tan t-1\right)\sqrt{\tan^2 t+1}\sec^2t\,dt\]
\[\frac{9}{4}\int\left(3\tan t-1\right)\sec^3t\,dt\]

Answer 8

Frankly I see no reason why the inverse sine function would be involved, unless there's a typo in the original integral. You have the ideal form for a tangent substitution.

Answer 9

Maybe you meant the inverse *hyperbolic* sine. That I can see...

Answer 10

idk cuz i can't figure out how this is even working

Answer 11

i have to input my answer online and it is asking for sin^ -1 so idk

Answer 12

it tells me to use # 30 in the ref page i just attached

Answer 13

#30 is intended for integrands of the form \(\sqrt{a^2-u^2}\), not \(\sqrt{a^2\color{red}+u^2}\).

Answer 14

idk why it is telling me that then. it says complete the square

Answer 15

\[I=\int\limits y \sqrt{10+4y+4y^2}dy\]
\[=\frac{ 1 }{ 8 } \int\limits \left( 8y \right)\sqrt{10+4y+4y^2}dy\]
\[=\frac{ 1 }{ 8 }\int\limits \left( 8y+4-4 \right)\sqrt{10+4y+4y^2}dy\]
\[=\frac{ 1 }{ 8 }\int\limits \left( 10+4y+4y^2 \right)^{\frac{ 1 }{ 2 }}\left( 8y+4 \right)dy-\frac{ 4 }{ 8 }\int\limits \sqrt{10+4y+4y^2}dy\]
\[=\frac{ 1 }{ 8 }\frac{ \left( 10+4y+4y^2 \right)^{\frac{ 3 }{ 2 }} }{ \frac{ 3 }{ 2 } }-\frac{ 1 }{ 2 }I _{1}+c\]

Answer 16

\[I _{1}=\int\limits \sqrt{10+4y+4y^2}dy=\int\limits \sqrt{4\left( y+\frac{ 1 }{ 2 } \right)^2+9}~dy=2~\int\limits \sqrt{\left( y+\frac{ 1 }{ 2 } \right) ^2+\frac{ 9 }{ 4 }} ~dy\]
=?