Integral formulas with proof

INTEGRATION

shahbaz ahmed


February 2024


√
Forms involving 2𝑎𝑥 − 𝑥2 , 𝑎>0




Introduction

𝑑𝑥
𝐼 =∫ √
2𝑎𝑥−𝑥2


2𝑎𝑥 − 𝑥2 = 𝑎2 − 𝑎2 + 2𝑎𝑥 − 𝑥2 = 𝑎2 − (𝑎2 − 2𝑎𝑥 + 𝑥2 ) = 𝑎2 − (𝑎 − 𝑥)2 = 𝑎2 − (𝑥 − 𝑎)2


Put 𝑥 − 𝑎 = 𝑎 sin 𝜃


2𝑎𝑥 − 𝑥2 = 𝑎2 − (𝑥 − 𝑎)2 = 𝑎2 − (𝑎 sin 𝜃)2 = 𝑎2 − 𝑎2 sin2 𝜃 = 𝑎2 (1 − sin2 𝜃) = 𝑎2 cos2 𝜃 Or


2𝑎𝑥 − 𝑥2 = 𝑎2 cos2 𝜃


Taking square root
√ √
2𝑎𝑥 − 𝑥2 = 𝑎2 cos2 𝜃 = 𝑎 cos 𝜃


Also 𝑥 − 𝑎 = 𝑎 sin 𝜃 Or



1

,𝑥 = 𝑎 + 𝑎 sin 𝜃 = 𝑎(1 + sin 𝜃)


Taking derivative with respect to 𝜃

𝑑[𝑎(1+sin 𝜃)]
𝑑𝑥
𝑑𝜃
= 𝑑𝜃
= 𝑎[ 𝑑[(1+sin
𝑑𝜃
𝜃)
] = 𝑎[ 𝑑(1)
𝑑𝜃
+ 𝑑(sin 𝜃)
𝑑𝜃
] = 𝑎[0 + cos 𝜃] = 𝑎 cos 𝜃 Or


𝑑𝑥 = 𝑎 cos 𝜃𝑑𝜃
√
Putting 2𝑎𝑥 − 𝑥2 = 𝑎 cos 𝜃 and 𝑑𝑥 = 𝑎 cos 𝜃𝑑𝜃 in the integral

𝑑𝑥 𝑎 cos 𝜃𝑑𝜃
𝐼 =∫ √ =∫ = ∫ 𝑑𝜃 = 𝜃 + 𝐶
2𝑎𝑥−𝑥2 𝑎 cos 𝜃


Now

𝑥−𝑎
𝑥 − 𝑎 = 𝑎 sin 𝜃 ⟹ 𝑎
= sin 𝜃 ⟹


sin−1 ( 𝑥−𝑎
𝑎
)=𝜃


Hence

𝑑𝑥
𝐼 =∫ √ = 𝜃 + 𝐶 = sin−1 ( 𝑥−𝑎
𝑎
)+𝐶
2𝑎𝑥−𝑥2


Now



𝑑𝑥
𝐼 =∫ √ = sin−1 ( 𝑥−𝑎
𝑎
)+𝐶
2𝑎𝑥−𝑥2



.....................................................



√
∫ 2𝑎𝑥 − 𝑥2 𝑑𝑥


2𝑎𝑥 − 𝑥2 = 𝑎2 − 𝑎2 + 2𝑎𝑥 − 𝑥2 = 𝑎2 − (𝑎2 − 2𝑎𝑥 + 𝑥2 ) = 𝑎2 − (𝑎 − 𝑥)2 = 𝑎2 − (𝑥 − 𝑎)2


Putting 𝑥 − 𝑎 = 𝑎 sin 𝜃


2

, 2𝑎𝑥 − 𝑥2 = 𝑎2 − (𝑥 − 𝑎)2 = 𝑎2 − (𝑎 sin 𝜃)2 = 𝑎2 − 𝑎2 sin2 𝜃 = 𝑎2 (1 − sin2 𝜃) = 𝑎2 cos2 𝜃


Taking square root
√ √
2𝑎𝑥 − 𝑥2 = 𝑎2 cos2 𝜃 = 𝑎 cos 𝜃
√ √
2𝑎𝑥 − 𝑥2 = 𝑎2 cos2 𝜃 = 𝑎 cos 𝜃 Also


𝑥 − 𝑎 = 𝑎 sin 𝜃 ⟹ 𝑥 = 𝑎 + 𝑎 sin 𝜃 = 𝑎(1 + sin 𝜃)


Taking derivative with respect to 𝜃

𝑑[𝑎(1+sin 𝜃]
𝑑𝑥
𝑑𝜃
= 𝑑𝜃
= 𝑎 𝑑(1+sin
𝑑𝜃
𝜃)
= 𝑎 cos 𝜃


𝑑𝑥 = 𝑎 cos 𝜃𝑑𝜃
√
Putting 2𝑎𝑥 − 𝑥2 = 𝑎 cos 𝜃 ,𝑑𝑥 = 𝑎 cos 𝜃𝑑𝜃 in the equation :
√
∫ 2𝑎𝑥 − 𝑥2 𝑑𝑥 = ∫ 𝑎 cos 𝜃𝑎 cos 𝜃𝑑𝜃 = ∫ 𝑎2 cos2 𝜃𝑑𝜃

1+cos 2𝜃
𝑐𝑜𝑠2 𝜃 = 2
⟹
√
∫ 2𝑎𝑥 − 𝑥2 𝑑𝑥 = ∫ 𝑎2 cos2 𝜃𝑑𝜃 = ∫ 𝑎2 [ 1+cos
2
2𝜃
𝑑𝜃]

𝑎2 𝑎2 𝑎2 𝑎2 𝑎2 sin 2𝜃 𝑎2 𝑎2 2 sin 𝜃 cos 𝜃
= 2
∫ [1 + cos 2𝜃]𝑑𝜃 = 2
∫ 𝑑𝜃 + 2
∫ cos 2𝜃𝑑𝜃 = 2
𝜃 + 2 2
+𝐶 = 2
𝜃 + 2 2

𝑎2 𝑎2
= 2
𝜃 + 2
sin 𝜃 cos 𝜃


Now

𝑥−𝑎
𝑥 − 𝑎 = 𝑎 sin 𝜃 ⟹ 𝑎
= sin 𝜃


Or ( 𝑥−𝑎
𝑎
)2 = sin2 𝜃 = 1 − cos2 𝜃 Or

(𝑥−𝑎)2 𝑎2 −(𝑥−𝑎)2 𝑎2 −𝑥2 −𝑎2 +2𝑎𝑥 2𝑎𝑥−𝑥2
cos2 𝜃 = 1 − 𝑎2
= 𝑎2
= 𝑎2
= 𝑎2
Or
√ √
2
cos2 𝜃 = 2𝑎𝑥−𝑥
𝑎2



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