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Question Number 175042 by peter frank last updated on 17/Aug/22
Answered by Frix last updated on 18/Aug/22
∫a2sin2x+b2cos2xa4sin4x+b4cos4xdx==∫(1+tan2x)(b2+a2tan2x)b4+a4tan4xdx==∫b2+a2tan2x(b4+a4tan4x)dtanx==∫a2t2+b2a4t4+b4dt==12∫dta2t2−ab2t+b2+12∫dta2t2+ab2t+b2==22ab(tan−1a2t−bb+tan−1a2t+bb)==−22abtan−1ab2ta2t2−b2==−22abtan−1ab2sinxcosxa2sin2x−b2cos2x+C
Commented by peter frank last updated on 18/Aug/22
thankyou
Commented by Tawa11 last updated on 20/Aug/22
Greatsir
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