Question 254 of 490MediumMCQ
If f′(x)=tan−1(secx+tanx), −π/2<x<π/2, f(0)=0, then f(1) equals: Solutions · 1
Official solution
\[\sec x+\tan x = \frac{1+\sin x}{\cos x} = \tan\left(\frac{\pi}{4}+\frac{x}{2}\right).\]
So f′(x)=4π+2x. Integrating: f(x)=4πx+4x2+C, C=0 (using f(0)=0).
\[f(1) = \frac{\pi}{4}+\frac14 = \frac{\pi+1}{4}.\] Answer the question to unlock