integrate 1/sin^4x+sin^2xcos^2x+cos^4x
Answers
solution:
I = ∫ dx/sin˄4x + cos˄4x – 2sin˄2xcos˄2x
= ∫dx/(sin˄2x+cos˄2x)˄2 – 2sin˄xcos˄2x +sin^2xcos˄2x
= ∫ dx/1-sin^2xcos^2x
= ∫ dx/1-sin^2x(1-sin^2x)
= ∫dx/1-sin^2x + sin^4x
= ∫ dx/cos^2x + sin^4x
Dividing numerator and denominator by cos^4x
= ∫ 1/cos^4x dx/cos2x/cos^4x + sin^4x/cos^4x
= ∫ sec^4x/sec^2x + tan^4x dx
= ∫sec^2x.sec^2x/sec^2x + tan^4x dx
= ∫ (1+ ten^2x)sec^2x/1+tan^2x + tan^4x dx
Putting tanx = t
Sec^2xdx = dt
Therefore,
I = ∫ 1+t^2/1+t^2 +t^4 dt
= ∫t^2 +1/t^4 + t^2 +1 dt
= ∫ t^2 (1+1/t^2)/t^2(t^2 + 1 + 1/t^2) dt
= ∫ (1+ 1/t^2)/t^2 +1/t^2 + 1) dt
= ∫ (1 + 1/t^2)/t – 1/t)^2 + (√3)^2 dt
Put, t – 1/t = u
( 1 + 1/t^2)dt = du
Therefore,
I = du/u^2 + (√3)^2
= 1/(√3) tan^ -1 (4/√3)+c
= 1/(√3) tan^ -1 (t – 1/t/√3) +c
= 1/(√3) tan^ -1 (t^2 – 1/√3) +c
= 1/(√3) tan^ -1 ( tan^2x -1/√3tanx) + c
Step-by-step explanation:
hope it will help plz give ur views. this is 5 marks question
