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1 ∫tan8×sec4dxevaluate the following2 ∫(4x-3)÷(x2+3x+8) dxevaluate

Prathibhaa P G , 7 Years ago
Grade 11
anser 1 Answers
Khimraj

Last Activity: 7 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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