Grade 11Integral Calculus
1 ∫tan8×sec4dxevaluate the following2 ∫(4x-3)÷(x2+3x+8) dxevaluate
1 ∫tan8×sec4dx
evaluate the following
2 ∫(4x-3)÷(x2+3x+8) dx
evaluate
1 Answer
Khimraj
I = ∫tan8×sec4xdx
I = ∫tan8×(1+tan2x)sec2xdx
Let tanx = t
then sec2xdx = dt
So I = ∫t8(1+t2)dt
I = ∫(t8+t10)dt
I = t9/9 + t11/11 +c
put t = tanx
So I = tan9x/9 + tan11x/11 +c.
Hope it clears. If you like answer then please approve it.
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