Integrals related "By Parts"!

Calculus Level 5

Consider the integrals: I 1 = ∫ 1 e ( 1 + x ) ( x + ln ⁡ x ) 100 d x I_{1}=\displaystyle \int_{1}^{e} (1+x)(x+\ln x)^{100} dx and I 2 = ∫ sin ⁡ − 1 1 e π 2 ( 1 + e sin ⁡ x + ln ⁡ sin ⁡ x ) 101 cos ⁡ x d x I_{2}=\displaystyle \int_{\sin^{-1}\frac{1}{e}}^{\frac{π}{2}} (1+e \sin x+\ln \sin x)^{101} \cos x \ dx

If I 1 + e I 2 101 = e ( 1 + e ) 101 − k 101 I_{1}+\dfrac{eI_2}{101} = \dfrac{e(1+e)^{101}-k}{101} , then find k k .


The answer is 1.

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1 solution

Patrick Corn
Mar 7, 2018

Let u = e sin ⁡ x , u = e\sin x, so ln ⁡ u = 1 + ln ⁡ sin ⁡ x \ln u = 1 + \ln \sin x and d u = e cos ⁡ x d x . du = e \cos x \, dx. Then I 1 + e 101 I 2 = I 1 + 1 101 ∫ 1 e ( u + ln ⁡ u ) 101 d u = I 1 + 1 101 ( u ( u + ln ⁡ u ) 101 ∣ 1 e − ∫ 1 e 101 u ( 1 + 1 / u ) ( u + ln ⁡ u ) 100 d u ) = I 1 + 1 101 ( e ( e + 1 ) 101 − 1 ) − ∫ 1 e ( u + 1 ) ( u + ln ⁡ u ) 100 d u = 1 101 ( e ( e + 1 ) 101 − 1 ) , \begin{aligned} I_1 + \frac{e}{101} I_2 &= I_1 + \frac1{101} \int_1^e (u+\ln u)^{101} \, du \\ &= I_1 + \frac1{101} \left( u(u+\ln u)^{101} \Big|_1^e - \int_1^e 101u(1+1/u)(u+\ln u)^{100} \, du \right) \\ &= I_1 + \frac1{101} \left( e(e+1)^{101} - 1 \right) - \int_1^e (u+1)(u+\ln u)^{100} \, du \\ &= \frac1{101} \left( e(e+1)^{101} - 1 \right), \end{aligned} so the answer is just 1 . \fbox{1}.

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