How do I solve this root integral with reduction formulae?

[nhrock3]

\intop_{-\pi/2}^{+\pi/2}\sqrt{(144sin^2t+6)cos^2t}dt
how to approach it?

 

[Mark44]

\intop_{-\pi/2}^{+\pi/2}\sqrt{(144sin^2t+6)cos^2t}dt
how to approach it?

Rewrite your integral as
\int_{-\pi/2}^{+\pi/2}\sqrt{144sin^2t+6}~~cos(t)dt

On the interval of your integration limits, sqrt(cos²(t)) = +cos(t).

Now you can use a simple substitution, u = sin(t). Then a trig substitution can be used.

 

[nhrock3]

ok i got
\int_{-1}^{1}\sqrt{144k^2+6}~~dk
\sqrt{6}\int_{-1}^{1}\sqrt{\frac{144}{6}k^2+1}~~dk
now i putk=\frac{\sqrt{6}}{12}tanz
dk=\frac{\sqrt{6}}{12}\frac{1}{cos^2z}dz
and after a manipulation i have
\int_{-1}^{1}\frac{\sqrt{6}}{12}\frac{1}{cos^3z}dz
what do to now i am stuck

 

[Mark44]

1/cos³(z) = sec³(k).

Your textbook probably shows an example of this integral.

 

[nhrock3]

un fortunatly i don't have such integral
it doesn't change much
sec x=1/(cos x)
how to approach it?

 

[dextercioby]

\int \frac{dx}{\cos^3 x} = \int \frac{d(\sin x)}{(1-\sin^2 x)^2} = ...

 

[Mark44]

un fortunatly i don't have such integral
it doesn't change much
sec x=1/(cos x)
how to approach it?

There's even a Wikipedia topic about it - http://en.wikipedia.org/wiki/Integral_of_secant_cubed

 

[HallsofIvy]

More generally, if you have either sine or cosine to an odd power, you can factor out one to go with the "dx" and change the rest to the other function:

\int sin^3(x)dx= \int (sin^2(x))(sins(x) dx)= \int (1- cos^2(x))(sin(x)dx)

Now let u= cos(x), so that du= -sin(x)dx and the integral becomes
-\int (1- u^2)du

Similarly,
\int \frac{dx}{cos^3(x)}= \int \frac{cos(x)dx}{cos^2(x)}= \int\frac{cos(x)dx}{1- sin^2(x)}
And, letting u= sin(x), so that du= cos(x)dx,
\int \frac{du}{1- u^2}
which can be integrated by "partial fractions".

 

[nhrock3]

what if i have even power?

 

[hunt_mat]

There is a wikipedia page all about the integrals on this stuff. Also have a look at reduction formulae and this will give you a good idea on what to do.

If I recall correctly though a poster did a huge post about this on your previous post which had exactly the same integral as you did here...