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Q15  Find the integrals of the functions  \tan ^ 3 2x \sec 2x

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\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x
                               \\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x

Therefore integration of \tan ^ 3 2x \sec 2x = 
                                                                      \\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\.....................(i)
Let assume 

\sec 2x = t
So, that  2\sec 2x.\tan 2x\ dx =dt
Now, the equation (i) becomes,

\\\Rightarrow \frac{1}{2}\int t^2\ dt-\frac{\sec 2x}{2}+C\\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C\\ =\frac{(\sec 2x)^3}{6}-\frac{\sec 2x}{2}+C

 

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manish

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