$$\int \sqrt{ 1 + 2 \tan x ( \tan x + \sec x )} dx$$
Please tell me the way of solving such questions. like what could i assume sec x or sec x tan x to be equal to?
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1Could you edit the parentheses to make it clearer what the integral is? – Paul Sundheim Aug 19 '14 at 19:00
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$\sec x=\frac{1}{\cos x}$ – idm Aug 19 '14 at 19:04
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$\bf{My\; Solution::}$ Let $\displaystyle I=\int \sqrt{1+2\tan x \left(\tan x+\sec x\right)}dx = \int \sqrt{1+2\tan^2 (x)+2\tan x\cdot \sec x}dx$
So $\displaystyle I = \int \sqrt{1+\tan^2 x+\tan^2 x+2\tan x\cdot \sec x}dx=\int \sqrt{\left(\tan x+\sec x\right)^2}dx$
So $\displaystyle I = \int \tan xdx+\int \sec xdx = \ln\left|\sec x\right|+\ln \left|\sec x+\tan x\right|+\mathbb{C} = \ln \left|\sec x\cdot \left(\sec x+\tan x\right)\right|+\mathbb{C}$
juantheron
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a little question: if we are rigorous, shouldn't it be $\sqrt{(\tan x+\sec x)^2}=\bigl|\tan x+\sec x\bigr|$? – cjferes Aug 19 '14 at 19:21