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int0^pix/(a^2cos^2x+b^2sin^2x)dx...

`int_0^pix/(a^2cos^2x+b^2sin^2x)dx`

Text Solution

Verified by Experts

Given:`int_0^pix/(a^2cos^2x+b^2sin^2x)dx`
`I=int_0^pix/(a^2cos^2x+b^2sin^2x)dx`
Now use the property ` f(a−x)=f(x)`
`2I=int_0^pi pi/(a^2cos^2x+b^2sin^2x)dx`
`2I=2pi int_0^pi 1/(a^2cos^2x+b^2sin^2x)dx`
Now use the property`(2a−x)=f(x)`
`I=pi int_0^(pi/2) (sec^2xsec^2xdx)/((a^2+b^2tan^2x)^2)`
Let ` b tanx=a tan t implies bsec^2xdx=asec^2tdt implies sec^2xdx=a/b(1+tan^2t)dt`
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