If f(x)=1cos2x1+tanx then its anti-derivative F(x) given that F(0)=4 is

f(x)=1cos2x1+tan x Anti derivation of f(x)=F(x)+c ⇒ ∫f(x) dx=F(x)+c ∫sec2x1+tan x=F(x)+c 21+tan x =F(x)+c x=0 ⇒ 21+0=F(0)+c 2=4+c -2=c ∴ 2 1+tan x=F(x)-2 2 1+tan x+1=F(x)

If f(x)=1cos2x1+tanx then its anti-derivative F(x) given that F(0)=4 is

f(x)=1cos2x1+tan x Anti derivation of f(x)=F(x)+c ⇒ ∫f(x) dx=F(x)+c ∫sec2x1+tan x=F(x)+c 21+tan x =F(x)+c x=0 ⇒ 21+0=F(0)+c 2=4+c -2=c ∴ 2 1+tan x=F(x)-2 2 1+tan x+1=F(x)

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If $f (x) = \frac{1}{\cos^{2} x \sqrt{1 + \tan x}}$ then its anti-derivative F(x) given that F(0)=4 is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

none of these

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

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \frac{1}{\cos^{2} x \sqrt{1 + \tan x}} Anti derivation of f (x) = F (x) + c \Rightarrow \int f (x) dx = F (x) + c \int \frac{\sec^{2} x}{\sqrt{1 + \tan x}} = F (x) + c 2 \sqrt{1 + \tan x} = F (x) + c x = 0 \Rightarrow 2 \sqrt{1 + 0} = F (0) + c 2 = 4 + c - 2 = c ∴ 2 \sqrt{1 + \tan x} = F (x) - 2 2 (\sqrt{1 + \tan x} + 1) = F (x)$