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Grade: 12
        
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4 months ago

Answers : (1)

Arun
21959 Points
							

Let I = Integ.dx/(4cos^2 x + 9sin^2 x)

= Integ.dx/{9cos ^2 x(4/9 + tan^2 x)}. ( By dividing and multiplying the denominator with 9cos^2 x)

= Integ. (1/9)sec^2 x dx/{(2/3)^2 + tan^2 x)

Let tan x = t => sec^2 x dx = dt

So I = (1/9) Integ. dt/{(2/3)^2 + t^2}

= (1/9){1/(2/3)} tan^(-1) {t/(2/3)

= (1/9)(3/2)tan^(-1) 3t/2

= (1/6)tan^(-1) 3tanx / 2

So limit 0 to π/2 I

= 1/6{tan^(-1) infinity - tan^(-1) 0}

= (1/6)( π/2 - 0) = π/12

4 months ago
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