-d x-sin x-cos x-2 equalsA- -1-2tanx-2-8- cB- 1-2tanx-2-8- cC- 1-2x-2-8- cD- -1-2x-2-8- -

The correct option is D 1/√(2)cot ( x/2+π/8 )+cI=dx/sin x cos x+√(2) =∫dx/√(2) (sin x. sin π/4 cos x cos π/4+1 ) =1/√(2)∫dx/1 cos ( x+π/4 ) =1/√(2)∫dx/1 ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫d x/sin x-co...

Question

$\int \frac{d x}{s i n x - c o s x + \sqrt{2}}$ equals

- \frac{1}{\sqrt{2}} t a n (\frac{x}{2} + \frac{π}{8}) + c

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\frac{1}{\sqrt{2}} t a n (\frac{x}{2} + \frac{π}{8}) + c

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\frac{1}{\sqrt{2}} c o t (\frac{x}{2} + \frac{π}{8}) + c

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- \frac{1}{\sqrt{2}} c o t (\frac{x}{2} + \frac{π}{8}) + c

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Solution

The correct option is D $- \frac{1}{\sqrt{2}} c o t (\frac{x}{2} + \frac{π}{8}) + c$
$I = \frac{d x}{s i n x - c o s x + \sqrt{2}} = \int \frac{d x}{\sqrt{2} (s i n x . s i n \frac{π}{4} - c o s x c o s \frac{π}{4} + 1)} = \frac{1}{\sqrt{2}} \int \frac{d x}{1 - c o s (x + \frac{π}{4})} = \frac{1}{\sqrt{2}} \int \frac{d x}{1 - c o s 2 (\frac{x}{2} + \frac{π}{8})} = \frac{1}{\sqrt{2}} \int \frac{d x}{2 s i n^{2} (\frac{x}{2} + \frac{π}{8})} = \frac{1}{2 \sqrt{2}} \int c o s e c^{2} (\frac{x}{2} + \frac{π}{8}) d x = \frac{1}{2 \sqrt{2}} \frac{- c o t (\frac{x}{2} + \frac{π}{8})}{\frac{1}{2}} + c = - \frac{1}{\sqrt{2}} c o t (\frac{x}{2} + \frac{π}{8}) + c$

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∫dxsin x−cos x+√2 equals

The correct option is D −1√2cot(x2+π8)+cI=dxsin x−cos x+√2=∫dx√2(sin x.sinπ4−cos x cosπ4+1)=1√2∫dx1−cos(x+π4)=1√2∫dx1−cos 2(x2+π8)=1√2∫dx2 sin2(x2+π8)=12√2∫cosec2(x2+π8)dx=12√2−cot(x2+π8)12+c=−1√2cot(x2+π8)+c

$\int \frac{d x}{s i n x - c o s x + \sqrt{2}}$ equals