If f(x)=1cos2⁡x1+tan⁡x then its anti-derivative F(x) ,F(0) = 4 is

# If $f\left(x\right)=\frac{1}{{\mathrm{cos}}^{2}x\sqrt{1+\mathrm{tan}x}}$ then its anti-derivative is

1. A

$\sqrt{1+\mathrm{tan}x}+4$

2. B

$\frac{2}{3}\left(1+\mathrm{tan}x{\right)}^{3/2}$

3. C

$2\left(\sqrt{1+\mathrm{tan}x}+1\right)$

4. D

none of these

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### Solution:

$\begin{array}{l}F\left(x\right)=\int \frac{\mathrm{d}x}{{\mathrm{cos}}^{2}x\sqrt{1+\mathrm{tan}x}}+C\\ =\int \frac{1}{\sqrt{1+t}}\mathrm{d}t+C\left(t=\mathrm{tan}x\right)\\ =2\left(\sqrt{1+t}\right)+C\\ =2\sqrt{1+\mathrm{tan}x}+C\end{array}$

Since $4=F\left(0\right)=2+C⇒C=2$ Hence

$F\left(x\right)=2\left(\sqrt{1+\mathrm{tan}x}+1\right)$

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