- 0 - -2 - x - x - co x dx-

The correct option is B π 4 Let I=∫ _ 0 ^π /2 √( x #160;) #160;√( x #160;) +√(cosec x #160;) #160; #160; dx Put x=x π2 I=∫ ...

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Byju's Answer

Standard XII

Mathematics

Circular Measurement of Angle

∫ 0 π /2 √ x ...

Question

$\int_{0}^{π / 2} \frac{\sqrt[n]{sec x}}{\sqrt[n]{sec x} + \sqrt[n]{c o sec x}} d x =$

\frac{π}{2}

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\frac{π}{3}

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\frac{π}{4}

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\frac{π}{6}

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Solution

The correct option is B $\frac{π}{4}$
Let $I = \int_{0}^{π / 2} \frac{\sqrt[n]{sec x}}{\sqrt[n]{sec x} + \sqrt[n]{cosec x}} d x$
Put $x = x - \frac{π}{2}$
$⟹ I = \int_{0}^{π / 2} \frac{\sqrt[n]{cosec x}}{\sqrt[n]{cosec x} + \sqrt[n]{sec x}} d x$
Adding both the integration, we get
$2 I = \int_{0}^{π / 2} \frac{\sqrt[n]{sec x} + \sqrt[n]{cosec x}}{\sqrt[n]{cosec x} + \sqrt[n]{sec x}} d x$
$= \int_{0}^{π / 2} d x$
$= \frac{π}{2}$
$⟹ I = \frac{π}{4}$

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∫π/20n√secxn√secx+n√cosecxdx=

The correct option is B π4Let I=∫π/20n√secxn√secx+n√cosec xdxPut x=x−π2⟹I=∫π/20n√cosec xn√cosec x+n√secxdxAdding both the integration, we get2I=∫π/20n√secx+n√cosec xn√cosec x+n√secxdx =∫π/20dx =π2⟹I=π4

$\int_{0}^{π / 2} \frac{\sqrt[n]{sec x}}{\sqrt[n]{sec x} + \sqrt[n]{c o sec x}} d x =$