The value of -4 n-4-sinx-cosx-dx- is equal to

The correct option is B 2√(2)n #160;I=∫ _ π #160; 4 #160;^ nπ π #160;/ 4 #160;| sin x+cos x #160; #160;| #160; dx #160;=∫ _ ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

The value of ...

Question

The value of $\int_{- \frac{π}{4}}^{n π - \frac{π}{4}} | sin x + cos x | d x$ , is equal to

2 \sqrt{2} n

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\sqrt{2} n

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\frac{1}{2 \sqrt{2}} n

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none of these

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Solution

The correct option is B $2 \sqrt{2} n$
$I = \int_{- \frac{π}{4}}^{n π - \frac{π}{4}} | sin x + cos x | d x$
$= \int_{- \frac{π}{4}}^{n π - \frac{π}{4}} \sqrt{2} ∣ ∣ ∣ sin (x + \frac{π}{4}) ∣ ∣ ∣ d x$
$= n \int_{- \frac{π}{4}}^{π - \frac{π}{4}} \sqrt{2} ∣ ∣ ∣ sin (x + \frac{π}{4}) ∣ ∣ ∣ d x$
$∵ ∣ ∣ ∣ sin (x + \frac{π}{4}) ∣ ∣ ∣$ is periodic with period $π$
$= \sqrt{2} n \int_{- \frac{π}{4}}^{\frac{3 π}{4}} sin (x + \frac{π}{4}) d x = \sqrt{2} n {(- cos (x + \frac{π}{4}))}_{- \frac{π}{4}}^{\frac{3 π}{4}} = 2 \sqrt{2} n$

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The value of ∫nπ−π4−π4|sinx+cosx|dx, is equal to

The correct option is B 2√2nI=∫nπ−π4−π4|sinx+cosx|dx=∫nπ−π4−π4√2∣∣∣sin(x+π4)∣∣∣dx=n∫π−π4−π4√2∣∣∣sin(x+π4)∣∣∣dx∵∣∣∣sin(x+π4)∣∣∣ is periodic with period π=√2n∫3π4−π4sin(x+π4)dx=√2n(−cos(x+π4))3π4−π4=2√2n

The value of $\int_{- \frac{π}{4}}^{n π - \frac{π}{4}} | sin x + cos x | d x$ , is equal to