Calc.PartFrac1

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\int {\frac  {dx}{\sin ^{2}(x)-\cos ^{2}(x)}}

This may look formidable, but it takes only a couple of steps to get this to a plain old partial fractions problem.

\int {\frac  {dx}{\sin ^{2}(x)-\cos ^{2}(x)}} =\int {\frac  {\sec ^{2}(x)\,dx}{\sec ^{2}(x)\left(\sin ^{2}(x)-\cos ^{2}(x)\right)}}
=\int {\frac  {\sec ^{2}(x)\,dx}{\tan ^{2}(x)-1}}
=\int {\frac  {du}{u^{2}-1}} (making substitution (u=\tan(x))
=\int {\frac  {1}{2}}\left({\frac  {1}{u-1}}-{\frac  {1}{u+1}}\right)\,du
={\frac  {1}{2}}\left(\ln \vert u-1\vert -\ln \vert u+1\vert \right)+C
={\frac  {1}{2}}\ln \left\vert {\frac  {u-1}{u+1}}\right\vert +C
={\frac  {1}{2}}\ln \left\vert {\frac  {\tan(x)-1}{\tan(x)+1}}\right\vert +C

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