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In = int0^(pi/4) tan^n x dx , then the v...

`I_n = int_0^(pi/4) tan^n` x dx , then the value of `n(l_(n-1) + I_(n+1))` is

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`inttan^(n)xdx=inttan^(n-2)x*tan^(2)xdx`
`=inttan^(n-20x(sec^(2)x-1)dx`
`=inttan^(n-2)x*sec^(2)xdx-inttan^(n-2)xdx`
`=inttan^(n-2)x.d(tanx)-intan^(n-2)xdx`
`=(tan^(n-1)x)/(n-2+1)-inttan^(n-2)xdx`
`:.inttan^(n)xdx=(tan^(n-1)x)/(n-1)-inttan^(n-2)xdx`
Now `l_(n)=int_(0)^((pi)/(4))tan^(n)xdx`
`=[(tan^(n-1)x)/(n-1)]_(0)^((pi)/(4))-int_(0)^((pi)/(4))tan^(n-2)xdx`
`=(1)/(n-1)-int_(0)^((pi)/(4))tan^(n-2)xdx`
`impliesl_(n+1)=(1)/(n)-int_(0)^((pi)/(4))tan^(n-1)xdx` (changing `n` to `n+1`)
`impliesl_(n+1)=(1)/(n)-l_(n-1)`
`impliesn(l_(n+1)+l_(n-1))=1`
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