Integral of ∫dx/(a^2+ 〖tan〗^2 (x))= ?

**sltvm2007** · May 26th 2012, 05:36 AM

Help please

∫dx/(a^2+ 〖tan〗^2 (x))= ?

Tanks

**Prove It** · May 26th 2012, 07:28 AM

Originally Posted by sltvm2007

Help please

∫dx/(a^2+ 〖tan〗^2 (x))= ?

Tanks

$\displaystyle \begin{align*} \int{\frac{dx}{a^2 + \tan^2{x}}} &= \int{\frac{\sec^2{x}\,dx}{\sec^2{x}\left(a^2 + \tan^2{x}\right)}} \\ &= \int{\frac{\sec^2{x}\,dx}{\left(1 + \tan^2{x}\right)\left(a^2 + \tan^2{x}\right)}} \\ &= \int{\frac{du}{\left(1 + u^2\right)\left(a^2 + u^2\right)}}\textrm{ after making the substitution } u = \tan{x} \implies du = \sec^2{x}\,dx \\ &= \int{\frac{1}{\left(a^2 - 1\right)\left(1 + u^2\right)} - \frac{1}{\left(a^2 - 1\right)\left(a^2 + u^2\right)}\,du} \\ &= \frac{1}{a^2 - 1}\int{\frac{1}{1 + u^2}- \frac{1}{a^2 + u^2}\,du} \\ &= \frac{1}{a^2 - 1}\left(\arctan{u} - \frac{1}{a}\arctan{\frac{u}{a}}\right) + C \\ &= \frac{1}{a^2 - 1}\left[\arctan{\left(\tan{x}\right)} - \frac{1}{a}\arctan{\left(\frac{\tan{x}}{a}\right)} \right] + C \\ &= \frac{1}{a^2 - 1}\left[x - \frac{1}{a}\arctan{\left(\frac{\tan{x}}{a}\right)} \right] + C \end{align*}$

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