-secx2- x-1-2 x2 d x-fx-c- where C is constant of integration and f0-0- then -0- fx d x is equal toA- 0B- ln 2C- ln 3D- ln 4

The correct option is C ln 3 ∫1/cosx ( 2+1/cosx )/ ( 1+2/cosx )^2dx=∫2cosx+1/cos x+2^2 ∫cosx(coxx+2)+sin^2x/(1+2secx)^2dx =∫cosx/cos x+2dx+∫sin^2x/(cosx+2)^2 ...

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Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

∫secx2+ x/1+2...

Question

$\int \frac{s e c x (2 + s e c x)}{(1 + 2 s e c x)^{2}} d x = f (x) + c$ , where C is constant of integration and f(0) = 0, then $\int_{0}^{π} f (x) d x$ is equal to

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ln 2

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ln 3

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ln 4

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Solution

The correct option is C
ln 3

$\int \frac{\frac{1}{c o s x} (2 + \frac{1}{c o s x})}{{(1 + \frac{2}{c o s x})}^{2}} d x = \int {\frac{2 c o s x + 1}{c o s x + 2}}^{2} \int \frac{c o s x (c o x x + 2) + s i n^{2} x}{(1 + 2 s e c x)^{2}} d x = \int \frac{c o s x}{c o s x + 2} d x + \int \frac{s i n^{2} x}{(c o s x + 2)^{2}} = \frac{1}{c o s x + 2} s i n x + \int \frac{- s i n^{2} x}{(c o s x + 2)^{2}} s i n x d x + \int \frac{s i n^{2} x}{(c o s x + 2)^{2}} d x = \frac{s i n x}{c o s x + 2} + C$

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∫secx(2+secx)(1+2secx)2dx=f(x)+c, where C is constant of integration and f(0) = 0, then ∫π0f(x)dx is equal to

The correct option is C ln 3 ∫1cosx(2+1cosx)(1+2cosx)2dx=∫2cosx+1cosx+22∫cosx(coxx+2)+sin2x(1+2secx)2dx=∫cosxcosx+2dx+∫sin2x(cosx+2)2=1cosx+2sinx+∫−sin2x(cosx+2)2sinxdx+∫sin2x(cosx+2)2dx=sinxcosx+2+C

$\int \frac{s e c x (2 + s e c x)}{(1 + 2 s e c x)^{2}} d x = f (x) + c$ , where C is constant of integration and f(0) = 0, then $\int_{0}^{π} f (x) d x$ is equal to