-20 sin3x-2 dx-cos3x-2-sin3x-2- -Roorkee 1989- BIT Ranchi 1989-

Let I=∫^π/2_0 sin^3x/2 dx/cos^3x/2+sin^3x/2 ⋯(i) =∫^π/2_0 sin^3/2 ( π/2 x )/cos^3/2 ( π/2 x )+sin^3/2 ( π/2 x )dx =∫^π/2_0 cos^3x/2 dx/sin^3x/2+cos^3x/2 ...

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Byju's Answer

Standard XIII

Mathematics

Area between Two Curves

∫π/20 sin3x/2...

Question

$\int_{0}^{\frac{π}{2}} \frac{s i n^{\frac{3 x}{2}} d x}{c o s^{\frac{3 x}{2}} + s i n^{\frac{3 x}{2}}} =$ [Roorkee 1989; BIT Ranchi 1989]

0

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π

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\frac{π}{2}

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$\frac{π}{4}$

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Solution

The correct option is D
$\frac{π}{4}$

$L e t I = \int_{0}^{\frac{π}{2}} \frac{s i n^{\frac{3 x}{2}} d x}{c o s^{\frac{3 x}{2}} + s i n^{\frac{3 x}{2}}} \dots (i)$
$= \int_{0}^{\frac{π}{2}} \frac{s i n^{\frac{3}{2}} (\frac{π}{2} - x)}{c o s^{\frac{3}{2}} (\frac{π}{2} - x) + s i n^{\frac{3}{2}} (\frac{π}{2} - x)} d x$
$= \int_{0}^{\frac{π}{2}} \frac{c o s^{\frac{3 x}{2}} d x}{s i n^{\frac{3 x}{2}} + c o s^{\frac{3 x}{2}}} \dots (i i)$
Adding (i) and (ii), we get $I = \frac{1}{2} \int_{0}^{\frac{π}{2}} 1 d x = \frac{1}{2} [x]_{0}^{\frac{π}{2}} = \frac{π}{4}$

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$\int_{0}^{\frac{π}{2}} \frac{s i n^{\frac{3 x}{2}} d x}{c o s^{\frac{3 x}{2}} + s i n^{\frac{3 x}{2}}} =$ [Roorkee 1989; BIT Ranchi 1989]

Q. Exhaustive set to values of x satisfying

| cos 3 x + sin 3 x | = | cos 3 x | + | sin 3 x |

[0, \frac{π}{2}]

is :

Q. The greatest value of

{sin}^{3} x + {cos}^{3} x

[0, \frac{π}{2}]

Q. Solve

\frac{{sin}^{3} \frac{x}{2} - {cos}^{3} \frac{x}{2}}{2 + sin x} = \frac{cos x}{3}

Q. Solve:

\frac{{sin}^{3} \frac{x}{2} - {cos}^{3} \frac{x}{2}}{2 + sin x} = \frac{cos x}{3}

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∫π20 sin3x2 dxcos3x2+sin3x2= [Roorkee 1989; BIT Ranchi 1989]

The correct option is D π4 Let I=∫π20sin3x2 dxcos3x2+sin3x2 ⋯(i) =∫π20sin32(π2−x)cos32(π2−x)+sin32(π2−x)dx =∫π20cos3x2 dxsin3x2+cos3x2 ⋯(ii) Adding (i) and (ii), we get I=12∫π20 1dx=12[x]π20=π4

$\int_{0}^{\frac{π}{2}} \frac{s i n^{\frac{3 x}{2}} d x}{c o s^{\frac{3 x}{2}} + s i n^{\frac{3 x}{2}}} =$ [Roorkee 1989; BIT Ranchi 1989]