Plz solve this sum

(tanⁿx + tan^n-2x)dx, and the limits are 0 to π/4

tanⁿx + tan^n-2x = tanⁿx(1 + 1/tan²x) = tanⁿx((tan²x + 1)/tan²x) = tan^n-2(sec²x)

(tanⁿx + tan^n-2x)dx = tan^n-2(sec²x) dx 

Put tanx = y

sec²xdx = dy, and the limits x = 0, y = 0 and x = π/4, y = 1.

y^n-2dy = y^n-1/(n-1) + c

Put the limits,

y^n-2dy = 1/(n-1), where n is an integer except 1, i.e. n N - {1}

Regards,

Team,

TopperLearning.