maeveoneill
Junior Member
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- Sep 24, 2005
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can someone please explain to me how to find the itnegral of tan^2x?
thank you!
thank you!
integral of tan^2x
- Thread starter maeveoneill
- Start date
Let's derive the reduction formula. Then you can find the integral of all
\(\displaystyle \L\\\int{tan^{n}(x)}dx\)
Rewrite as:
\(\displaystyle \L\\\int{tan^{n-2}tan^{2}(x)}dx\)
=\(\displaystyle \L\\\int{tan^{n-2}(sec^{2}(x)-1)}dx\)
=\(\displaystyle \L\\\int{tan^{n-2}(x)sec^{2}(x)}dx-\int{tan^{n-2}(x)}dx\)
The first integral above can be evaluated using and
\(\displaystyle \L\\\int{tan^{2}(x)}dx=\int{u^{n-2}}du-\int{tan^{n-2}(x)}dx\)
=\(\displaystyle \L\\\frac{u^{n-1}}{n-1}-\int{tan^{n-2}(x)}dx\)
=\(\displaystyle \L\\\frac{1}{n-1}tan^{n-1}(x)-\int{tan^{n-2}(x)}dx\)
Which is the reduction formula for
If n=2, then the integral part of this is
\(\displaystyle \L\\\int{tan^{n}(x)}dx\)
Rewrite as:
\(\displaystyle \L\\\int{tan^{n-2}tan^{2}(x)}dx\)
=\(\displaystyle \L\\\int{tan^{n-2}(sec^{2}(x)-1)}dx\)
=\(\displaystyle \L\\\int{tan^{n-2}(x)sec^{2}(x)}dx-\int{tan^{n-2}(x)}dx\)
The first integral above can be evaluated using and
\(\displaystyle \L\\\int{tan^{2}(x)}dx=\int{u^{n-2}}du-\int{tan^{n-2}(x)}dx\)
=\(\displaystyle \L\\\frac{u^{n-1}}{n-1}-\int{tan^{n-2}(x)}dx\)
=\(\displaystyle \L\\\frac{1}{n-1}tan^{n-1}(x)-\int{tan^{n-2}(x)}dx\)
Which is the reduction formula for
If n=2, then the integral part of this is