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Question

Integrate the function  $$\sec^2(7-4x)$$


Solution

Let $$7-4x=t$$
$$\therefore -4dx=dt$$
$$\therefore\displaystyle  \int \sec^2(7-4x)dx=\frac {-1}{4}\int sec^2t dt$$
$$\displaystyle =\frac {-1}{4}(\tan t)+C$$
$$\displaystyle =\frac {-1}{4}\tan (7-4x)+C$$

Mathematics
RS Agarwal
Standard XII

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