If -1- x - cosec x - tan x - x2dx - x- - cos x - -4- - cos 2x- - c- then - - - -2- - - is - where c is an arbitrary constant

∫ (sin x cos x/1 + sin x + cos x)^2 dx = 1/4∫ (2 sin x cos x/1 + sin x + cos x)^2 dx = 1/4∫ ((sin x + cos x)^2 1/sin x + cos x + 1)^2 dx = 1/4∫(sin x + cos ...

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Byju's Answer

Standard XIII

Mathematics

Integration of Trigonometric Functions

If ∫1/ x + co...

Question

$If \int \frac{1}{(sec x + cosec x + tan x + cot x)^{2}} d x = \frac{x}{α} + \frac{cos (x + \frac{π}{4})}{β} + \frac{cos 2 x}{γ} + c, then | α + \sqrt{2} β + γ | is$ , (where $c$ is an arbitrary constant)

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Solution

$\int (\frac{sin x cos x}{1 + sin x + cos x})^{2} d x = \frac{1}{4} \int (\frac{2 sin x cos x}{1 + sin x + cos x})^{2} d x = \frac{1}{4} \int (\frac{(sin x + cos x)^{2} - 1}{sin x + cos x + 1})^{2} d x = \frac{1}{4} \int \frac{(sin x + cos x + 1)^{2} (sin x + cos x - 1)^{2}}{(1 + sin x + cos x)^{2}} = \frac{1}{4} \int (2 + 2 sin x cos x - 2 sin x - 2 cos x) d x = \frac{x}{2} - \frac{cos 2 x}{8} + \frac{1}{2} (cos x - sin x) + c = \frac{x}{2} - \frac{cos 2 x}{8} + \frac{cos (x + \frac{π}{4})}{\sqrt{2}} + c α = 2, β = \sqrt{2}, γ = - 8 | α + \sqrt{2} β + γ | = | 4 - 8 | = 4$

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If ∫1(sec x+cosec x+tan x+cot x)2dx=xα+cos(x+π4)β+cos 2xγ+c,then |α+√2β+γ| is , (where c is an arbitrary constant)

$If \int \frac{1}{(sec x + cosec x + tan x + cot x)^{2}} d x = \frac{x}{α} + \frac{cos (x + \frac{π}{4})}{β} + \frac{cos 2 x}{γ} + c, then | α + \sqrt{2} β + γ | is$ , (where $c$ is an arbitrary constant)