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one month ago

Arun
20410 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 = dtSo 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 / 2So limit 0 to π/2 I= 1/6{tan^(-1) infinity - tan^(-1) 0}= (1/6)( π/2 - 0) = π/12
```
one month ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions