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Letting x=π2−x and using the property that
∫a0f(x)dx=∫a0f(a−x)dx
we obtain
∫π20tanx(1+(π2−x)2)(1+tanx)dx
Now, add equation (1) and (2). After that I do not understand how I can proceed fυятнєя
ι нσρє уσυ нєℓρ
∫a0f(x)dx=∫a0f(a−x)dx
we obtain
∫π20tanx(1+(π2−x)2)(1+tanx)dx
Now, add equation (1) and (2). After that I do not understand how I can proceed fυятнєя
ι нσρє уσυ нєℓρ
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