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प्रश्न
Evaluate : `∫(sin^6x+cos^6x)/(sin^2x.cos^2x)dx`
उत्तर
`int(sin^6x+cos^6x)/(sin^2x.cos^2x)dx`
`=int((sin^2x+cos^x)^2-3sin^2x.cos^2x(sin^2x+cos^2x))/(sin^2x.cos^2x)dx [Using a^3+b^3=(a+b)^3−3ab(a+b)]`
`=int(1-3sin^2x.cos^2x)/(sin^2xcos^2x)dx [Using sin^2x+cos^2x=1]`
`=int(1/(sin^2x.cos^2x)-3)dx`
`=int((sin^2x+cos^2x)/(sin^2x.cos^2x)-3)dx`
`=int(sec^2x+cosec^2x-3)dx`
`=intsec^2xdx+intcosec^2xdx-int3dx`
`=tanx-cotx-3x+C`
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