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प्रश्न
Evaluate : `int (sinx)/sqrt(36-cos^2x)dx`
उत्तर
`intsinx/sqrt(36-cos^2x)dx`
Substitute, cosx = t
∴ - sin x dx = dt
∴ sin x dx = - dt
The integral becomes
`int (-dt)/sqrt( 36 - t^2 )`
= `-intdt/sqrt( 6^2 - t^2 )`
= `-sin^-1( t/6 ) + C`
= `-sin^-1 .(cosx/6) + c`
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