Definite Integral

Question

$\displaystyle S = \int_{0}^{2\pi } \sqrt{4\cos^{2}t +\sin^{2}t} \, \, dt$

Please explain how to solve it.

Comments

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you can reach at [email protected]

Answer 1

$\int_{0}^{2\Pi }\sqrt{4\cos^{2}t +sin^{2}t}dt$

$=\int_{0}^{2\Pi }\sqrt{3\cos^{2}t +1}dt$

$=\sqrt{3}\int_{0}^{2\Pi }\sqrt{\cos^{2}t +\left ( \frac{1}{\sqrt{3}} \right )^{2}}dt$

Now put in direct formula

$=\frac{cost\sqrt{cos^{2}t+\frac{1}{3}}}{2}+\frac{1/3}{2}log\begin{vmatrix} cos^{2}t & +&\sqrt{cos^{2}t+\frac{1}{3}} \end{vmatrix}$

As we know, $\int \sqrt{x^{2}+a^{2}}dx=\frac{x\sqrt{x^{2}+a^{2}}}{2}+\frac{a^{2}}{2}log\begin{vmatrix} x &+ & \sqrt{x^{2}+a^{2}} \end{vmatrix}$

Now put the range on it

Comments

@srestha , sorry I am unable to solve -$\int \frac{1}{u^{2}}\sqrt{\frac{u^{2} + 3}{u^{2}-1}} du$ . if possible then please show the steps to solve this integral or if you know any other method to solve $\int_{0}^{\Pi } \sqrt{4cos^{2}t + sin^{2}t} \, \, dt$ then please explain the procedure.

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http://www.wolframalpha.com/input/?i=+root+of+3+cos+%5E2+x%2B1

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http://www.wolframalpha.com/input/?i=+root+of+3+cos+%5E2+x%2B1

:o