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Q.

The value of the integral ∫0π/2 dx1+16sin2⁡ x is

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a

π267

b

π323

c

π6

d

π2

answer is A.

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Detailed Solution

I=∫0π/2 dx1+16sin2⁡ x=∫0π/2 sec2⁡ xdxsec2⁡ x+16tan2⁡ x=67∫0∞ dt6/7+t2(t=tan⁡ x)=67×76tan−1⁡ t760∞=67⋅π2.
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The value of the integral ∫0π/2 dx1+16sin2⁡ x is