Integration problem

[mattflint50]

When I was studying for my calculus examine I came accross a problem that asked me do integrate tan^4 xdx. I tried everything imagineable butI cannot get it to come out. I tried U substitution and I foudn that that couldnt work. SO then I turned tan^4 into (sec^2x-1)^2. I then foiled this to get Sec^4x-2Sec^2x+1. Am I doing this right, if so where do I go from here.

 

[pka]

\(\displaystyle \L
\tan ^4 (x) = \tan ^2 (x)\left( {1 + \sec ^2 (x)} \right) = \left( {\tan ^2 (x) + \tan ^2 (x)\sec ^2 (x)} \right)\)

 

[galactus]

General formula:

\(\displaystyle \L\\\int{tan^{n}(x)dx=\frac{tan^{n-1}(x)}{n-1}-\int{tan^{n-2}(x)dx)\)


\(\displaystyle \L\\\int{tan^{4}(x)dx\)=\(\displaystyle \L\\\int{tan^{2}(x)tan^{2}(x)}dx\)

=\(\displaystyle \L\\\int{tan^{2}(x)(sec^{2}(x)-1)}dx\)=

\(\displaystyle \L\\\int{tan^{2}(x)sec^{2}(x)dx-\int{tan^{2}(x)}dx\)

=\(\displaystyle \L\\\frac{tan^{3}(x)}{3}-tan(x)+x+C\)

 

[mattflint50]

what do I do with the Tanx^2 Sec^2. Do I do U substitution

 

[mattflint50]

Ok thank you very much, I understand now

 

[galactus]

what do I do with the Tanx^2 Sec^2. Do I do U substitution

Let u=tan(x), then du=sec^2(x)