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

Can you answer this question?

Toolbox:
• (i)Method of substitution :
• Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
• Consider $I=\int f(x)dx.$
• Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
• $\Rightarrow$dx=g'(t)dt.
• Thus $I=\int f(g(t).g'(t))dt.$
• (ii)$\int \sec^2x=\tan x+c.$
Given $I=\int \sec^2(7-4x).$

Put 7-4x=t.

dt=-4dx $\Rightarrow dx=\frac{-dt}{4}.$

Now substituting for x and dx we get,

$I=\int \sec^2t\big(\frac{-dt}{4}\big).$

$\;\;\;=\frac{-1}{4}\int\sec^2tdt$.

On integrating we get,

$\frac{-1}{4}\tan t+c.$

Substituting back for t we get,

$\int\sec^2(7-4x)=\frac{1}{4}\tan(7-4x)+c.$

answered Jan 29, 2013