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If f(x)=∫ex(tan⁡x+tan⁡2x) dxf(x)=\int e^x(\tan x+\tan^2 x)\,dx and f(0)=0f(0)=0, then f(π4)f\left(\dfrac{\pi}{4}\right) equals:

Solution

✅ Correct Option: 3

Write tan⁡x+tan⁡2x=(tan⁡x−1)+sec⁡2x\tan x + \tan^2x = (\tan x - 1) + \sec^2 x, which is of the form g(x)+g′(x)g(x) + g'(x) with g(x)=tan⁡x−1g(x) = \tan x - 1.

So f(x)=ex(tan⁡x−1)+Cf(x) = e^x(\tan x - 1) + C. From f(0)=−1+C=0f(0) = -1 + C = 0, we get C=1C = 1.

Then f(π/4)=eπ/4(1−1)+1=1f(\pi/4) = e^{\pi/4}(1-1) + 1 = 1.

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