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Integrate the following functions: tan^4x |
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Answer» Solution :`tan^4x = tan^2 X tan^2 x` =`tan^2 x sec^2 x-(sec^2 x-1)` =`tan^2 x sec^2 x -sec^2 x+1` THEREFORE `INT tan^4 x dx` =`int (ta_n^2x sec^2 x-sec^2 x+1)dx` `int t^2dt-TANX+x+c`, if we PUT tanx =t =`t^3/3-tanx+x+c` =`tan^3 x/3 -tanx+x+c` |
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