Question
Question: Let f(0) = f ' (0) = 0 and f " (x) = sec<sup>4</sup>x + 4, then f(x) =...
Let f(0) = f ' (0) = 0 and f " (x) = sec4x + 4, then f(x) =
A
32log sec x + 61 tan2 x + 2x2
B
32log cos x + 61tan2 x + 2x2
C
32log sec x + 61 tan2 x
D
None of these
Answer
32log sec x + 61 tan2 x + 2x2
Explanation
Solution
Integrate we have
f ' (x) = ∫sec2x.(1+tan2x)dx+ 4x + C1
f ' (x) = tan x + 3tan3x + 4x + c1
put x = 0
a = 0 \end{matrix}$$ again integrate f(x) = log sec x + $\frac{1}{3}$ $\int_{}^{}{\tan^{3}xdx}$ + 2x<sup>2</sup> + c<sub>2</sub> = log secx + $\frac{1}{3}$ $\frac{\tan^{2}x}{2}$ – $\frac{1}{3}$ log sec x + c<sub>2</sub> + 2n<sup>2</sup>put x = 0 c<sub>2</sub> = 0 \\ $\begin{matrix} f(x) = 2x^{2} + \frac{1}{6}\tan^{2}x + \frac{2}{3}{logsec}x \end{matrix}$