-cos x-3 d x-1-4 sinx-3-4 sin 2x-3-A- cos x-1-2 sinx-3-cB- x-1-2 sinx-3-cC- 1-2tan -11-2 sin x -3- -D- sin x-1-2 sinx-3-c

The correct option is D sin x/1+2sin(x+ π/3) + cI = ∫(cos x + √(3))dx/(1+2 sin(x+ π/3))^2 = ∫cos x + √(3)/(1+sin x + √(3)cos x)^2 dx = ∫ x x + √(3)^2 x/( x + ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

∫cos x+√3 d x...

Question

$\int \frac{(cos x + \sqrt{3}) d x}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} =$

\frac{cos x}{1 + 2 sin (x + \frac{π}{3})} + c

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\frac{sec x}{1 + 2 sin (x + \frac{π}{3})} + c

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\frac{sin x}{1 + 2 sin (x + \frac{π}{3})} + c

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\frac{1}{2} {tan}^{- 1} (1 + 2 sin (x + \frac{π}{3})) + c

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Solution

The correct option is C $\frac{sin x}{1 + 2 sin (x + \frac{π}{3})} + c$
$I = \int \frac{(cos x + \sqrt{3}) d x}{(1 + 2 sin (x + \frac{π}{3}))^{2}}$
$= \int \frac{cos x + \sqrt{3}}{(1 + sin x + \sqrt{3} cos x)^{2}} d x$

$= \int \frac{csc x cot x + \sqrt{3} {csc}^{2} x}{(csc x + 1 + \sqrt{3} cot x)^{2}} d x$ {Dividing by ${sin}^{2} x$ }

$= \frac{1}{csc x + 1 + \sqrt{3} cot x} + c$ {let $csc x + 1 + \sqrt{3} cot x = t$ }

$= \frac{sin x}{1 + 2 sin (x + \frac{π}{3})} + c$

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Similar questions

\int \frac{cos x + \sqrt{3}}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} d x

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∫(cosx+√3)dx1+4sin(x+π3)+4sin2(x+π3)=

The correct option is C sinx1+2sin(x+π3)+cI=∫(cosx+√3)dx(1+2sin(x+π3))2 =∫cosx+√3(1+sinx+√3cosx)2dx =∫cscxcotx+√3csc2x(cscx+1+√3cotx)2dx {Dividing by sin2x} =1cscx+1+√3cotx+c {let cscx+1+√3cotx=t} =sinx1+2sin(x+π3)+c

$\int \frac{(cos x + \sqrt{3}) d x}{1 + 4 sin (x + \frac{π}{3}) + 4 {sin}^{2} (x + \frac{π}{3})} =$