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Integration of sin^6x/cos^8x dx
Please solve this question fast

Bhavit Jain , 6 Years ago
Grade 12
anser 1 Answers
Aditya Gupta

Last Activity: 6 Years ago

this is very ez to do.
sin^6x/cos^8x= (sin^6x/cos^6x)*sec^2x= tan^6x*sec^2x
now substitute tanx=y or dy= sec^2x dx
so that ∫sin^6x/cos^8x dx = ∫tan^6x*sec^2x dx= ∫y^6 dy= y^7/7+C
tan^7x/7 + C

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