Finding $\int 2t\sqrt{4\sin^2(t) + 9\cos^2(t) + 4}\,dt$

Question

I'm trying to find the result of the integral above.

I have tried some simple substitution, and got it down to:

$\int 2t\sqrt{8 + 5\cos^2(t)}\,dt$, which looks innocent enough. The solution is eluding me, however.

I have obtained this problem through my university, however, it is not worth any fraction of the grade. It is thought that such integrals are easy, but I am having difficulty. If someone could point me in the right direction, that'd be great.

Answer 1

Hint: Integrating by parts, we arrive at the conclusion that evaluating this expression is equivalent to finding the anti-derivative of an elliptic integral E of argument $\sqrt{\dfrac5{13}}.~$ However, this is not known to possess a closed form, except perhaps in terms of $($ generalized $)$ hypergeometric functions.

Answer 2

$$\int 2t\sqrt{8 + 5\cos^2t}\ dt$$ In an attempt to apply integration by parts, we have $$u=2t \Rightarrow du=2\ dt$$ $$v=\int\sqrt{8 + 5\cos^2t}\ dt$$ $$=\int\sqrt{13-5\sin^2t}\ dt$$ $$=\sqrt{13}\int\sqrt{1-\frac{5}{13}\sin^2t}\ dt$$ $$=\sqrt{13}E\left(t\ \bigg|\frac{5}{13}\right)+C$$ So now we end up with $$2\sqrt{13}tE\left(t\ \bigg|\frac{5}{13}\right)-2\sqrt{13}\int E\left(t\ \bigg|\frac{5}{13}\right)dt$$ Where $E$ is the incomplete elliptic integral of the second kind. I strongly doubt that this has a closed form in terms of elementary functions.

Answer 3

$\int2t\sqrt{4\sin^2t+9\cos^2t+4}~dt$

$=\int2t\sqrt{4\sin^2t+9(1-\sin^2t)+4}~dt$

$=\int2t\sqrt{13-5\sin^2t}~dt$

$=\int2\sqrt{13}t\sqrt{1-\dfrac{5\sin^2t}{13}}~dt$

$=\int\left(2\sqrt{13}t+\sum\limits_{n=1}^\infty\dfrac{5^n(2n)!t\sin^{2n}t}{2^{2n-1}13^{n-\frac{1}{2}}(n!)^2(1-2n)}\right)dt$

For $n$ is any natural number,

$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts.

$\therefore\int\left(2\sqrt{13}t+\sum\limits_{n=1}^\infty\dfrac{5^n(2n)!t\sin^{2n}t}{2^{2n-1}13^{n-\frac{1}{2}}(n!)^2(1-2n)}\right)dt$

$=\sqrt{13}t^2+\sum\limits_{n=1}^\infty\int\dfrac{5^n(2n)!t}{2^{2n-1}13^{n-\frac{1}{2}}(n!)^2(1-2n)}d\left(\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}\right)+C$

$=\sqrt{13}t^2+\sum\limits_{n=1}^\infty\dfrac{5^n((2n)!)^2t^2}{2^{4n-1}13^{n-\frac{1}{2}}(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2t\sin^{2k-1}t\cos t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k-1)!(1-2n)}-\sum\limits_{n=1}^\infty\int\left(\dfrac{5^n((2n)!)^2t}{2^{4n-1}13^{n-\frac{1}{2}}(n!)^4(1-2n)}-\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2\sin^{2k-1}t\cos t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k-1)!(1-2n)}\right)dt+C$

$=\sqrt{13}t^2+\sum\limits_{n=1}^\infty\dfrac{5^n((2n)!)^2t^2}{2^{4n-1}13^{n-\frac{1}{2}}(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2t\sin^{2k-1}t\cos t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k-1)!(1-2n)}-\sum\limits_{n=1}^\infty\dfrac{5^n((2n)!)^2t^2}{2^{4n}13^{n-\frac{1}{2}}(n!)^4(1-2n)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2\sin^{2k}t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k)!(1-2n)}+C$

$=\sqrt{13}t^2-\sum\limits_{n=1}^\infty\dfrac{5^n((2n)!)^2t^2}{2^{4n-1}13^{n-\frac{1}{2}}(n!)^4(1-2n)}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2t\sin^{2k-1}t\cos t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k-1)!(1-2n)}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{5^n((2n)!)^2((k-1)!)^2\sin^{2k}t}{2^{4n-2k+1}13^{n-\frac{1}{2}}(n!)^4(2k)!(1-2n)}+C$