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1 ∫tan 8 ×sec 4 dx evaluate the following 2 ∫(4x-3)÷(x 2 +3x+8) dx evaluate

1 ∫tan8×sec4dx
evaluate the following
2 ∫(4x-3)÷(x2+3x+8) dx
evaluate

Grade:11

1 Answers

Khimraj
3007 Points
3 years ago
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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