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Ans)
You can integrate tanθ
write tanθ=sinθ/cosθso now, integral of tanθ.dθ=integral of sinθ/cosθ.dθ→1put cosθ=t→2there fore -sinθ.dθ=dt→3put 2 and 3 in 1so,the question becomes integral of tanθ.dθ=integral of -dt/t→4=-log t +c → 5 [reason:dx/x=logx]put 2 in 5, =-log cosθ+Cwhere C is the integration constant
4.2^2x+1-9.2^x+1=0
how do you slove 4u-5=2u-13
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Example Multiply 3x 5 + 4x 3 + 2x - 1 and x 4 + 2x 2 + 4. The product is given by 3x 5 . (x 4 + 2x 2 + 4) + 4x 3 . (x 4 + 2x 2 + 4) + 2x .
yes you can intergrate tan θ tan q = sin q/cos q (1) let, y = cos q so, dy/dq = -sin q this implies, dq=dy/-sin q (2) putting values in (1) ∫ 1/-y = -log|y| =-log|cos q| so the ∫ tan q = -log|cos q|
yes you can intergrate tan θ
tan q = sin q/cos q (1)
let,
y = cos q
so, dy/dq = -sin q
this implies, dq=dy/-sin q (2)
putting values in (1)
∫ 1/-y
= -log|y|
=-log|cos q|
so the ∫ tan q = -log|cos q|
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