Question

Integrate dx/ 3sin^2x+5cos^2x

Answer 1

Given : ∫ dx/(3sin²x + 5cos²x)

To Find : Integrate

Solution:

∫ dx/(3sin²x + 5cos²x)

Divide numerator & denominator by cos²x

1/cos²x = sec² x   , sin²x/cos²x = tan² x  

= ∫ sec²x dx / ( 3tan²x + 5)

= (1/3)∫ sec²x dx / (  tan²x + 5/3)  .   . Eq1

Assume tanx = y

sec²x dx  = dy

Substitute in Eq1

=  (1/3)∫  dy / (  y²  + 5/3)

= (1/3)∫  dy / (  y²  + (√5/√3)²)

∫ dx/(x² + a²)  = (1/a)tan⁻¹(x/a) + c

= (1/3) (1/√5/√3) tan⁻¹( y/(√5/√3) + c

= (1/√15)tan⁻¹(√3y/√5) + c

y =  tanx

=  (1/√15)tan⁻¹(√3tanx/√5) + c

∫ dx/(3sin²x + 5cos²x) =  (1/√15)tan⁻¹(√3tanx/√5) + c  

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