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pushpa_bhagat · Feb 19, 2011, 08:54 AM

all the sin & cos terms are in denomenater .

galactus · Feb 19, 2011, 09:34 AM

$\int\frac{1}{(sin\theta-2cos\theta)(2sin\theta+cos\theta)}d\theta$

Sometimes these tough ones can be done by using obscure-looking identities. Rewrite as:

*** $\frac{-2}{5}\int\frac{1}{cos(2\theta-tan^{-1}(\frac{3}{4}))}d\theta$

Let $u=2\theta-tan^{-1}(\frac{3}{4}), \;\ \theta=\frac{u+tan^{-1}(\frac{3}{4})}{2}, \;\ d\theta=\frac{du}{2}$

Making the subs, the integral becomes:

$\frac{-1}{5}\int sec(u)du$

$=\frac{-1}{5}ln(sec(u)+tan(u))$

Resub:

$\frac{-1}{5}ln\left[sec\left(2\theta-tan^{-1}\left(\frac{3}{4}\right)\right)+tan\left(2\theta-tan^{-1}\left(\frac{3}{4}\right)\right)\right]+C$

***This can be derived by using the subtraction formula for cosine.

By the way, $\theta$ is spelled 'theta', not thita.