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प्रश्न
Evaluate `int_0^(pi/2) (tan^7x)/(cot^7x + tan^7x) "d"x`
उत्तर
We have I = `int_0^(pi/2) (tan^7x)/(cot^7x + tan^7x) "d"x` ....(1)
= `int_0^(pi/2) (tan^7(pi/2 - x))/(cot^7(pi/2 - x) + tan^7(pi/2 - x)) "d"x` ......By (p4)
= `int_0^(pi/2) (cot^7 (x) "d"x)/(cot^7x "d"x + tan^7x)` .....(2)
Adding (1) and (2), we get
2I = `int_0^(pi/2) ((tan^7x + cot^7x)/(tan^7x + cot^7x))"d"x`
= `int_0^(pi/2) "d"x` which gives I = `pi/4`
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