Question

Evaluate : `int (sinx)/sqrt(36-cos^2x)dx`

Answer 1

`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`