Using the substitution `x=tan^(2)theta`, evaluate
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx` (4 marks)
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Using the substitution `x=tan^(2)theta`, evaluate
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx` (4 marks)
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`pi/2-1`
`x=tan^(2)theta`
`dx/(d theta)=2sec^2theta\ tantheta\ \ =>\ \ dx=2sec^2theta\ tantheta\ d theta`
`text{When}\ x=0, \ theta=0`
`text{When}\ x=1, \ theta=pi/4`
| `sin^(-1)sqrt((x)/(1+x))` | `=sin^(-1)sqrt((tan^(2)theta)/(1+tan^(2)theta))` | |
| `=sin^(-1)sqrt((tan^(2)theta)/(text{sec}^(2)theta))` | ||
| `=sin^(-1)sqrt((sin^(2)theta))` | ||
| `=sin^(-1)(sintheta)` | ||
| `=theta` |
`int_(0)^(1)sin^(-1)sqrt((x)/(1+x))\ dx=int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
`text{Integrating by parts:}`
| `u` | `=theta` | `u^{′}` | `=1` |
| `v^{′}` | `=2tan theta\ sec^2 theta` | `v` | `=tan^(2)theta` |
`int_(0)^(pi/4)\ theta xx 2sec^2theta\ tantheta\ d theta`
`=[theta\ tan^(2)theta]_0^(pi/4)-int_0^(pi/4)tan^(2)theta\ d theta`
`=(pi/4 xx 1 -0)-int sec^(2)theta-1\ d theta`
`=pi/4-[tan theta-theta]_0^(pi/4)`
`=pi/4-[(1-pi/4)-0]`
`=pi/2-1`