Calculus Integration Problems, part II?

[匿名]

2013-05-02 3:39 pm

5. Integrate ∫ 1 / (x^2 + 2x + 6) dx

6. If y = arcsin [e^(x^2)], find dy/dx.

7. Integrate ∫ cox^3 x sin^25 x dx

Could you kindly show work with exact steps? I am not very good with integration by parts. Thanks!

回答 (1)

An ESL Learner

2013-05-02 5:26 pm

✔ 最佳答案

5)
∫ 1/(x^2 + 2x + 6) dx
= ∫ 1/((x + 1)^2 + (√5)^2) dx

Let tanθ = (x + 1)/√5.
x = √5tanθ - 1
dx = √5sec^2(θ) dθ

secθ = √((x + 1)^2 + (√5)^2)/√5
(x + 1)^2 + (√5)^2 = 5sec^2θ

= ∫ 1/(5sec^2(θ)) * √5sec^2(θ) dθ
= ∫ 1/√5 dθ
= θ/√5 + C
= arctan((x + 1)/5)/√5 + C

6)
y = arcsin[e^(x^2)]
siny = e^(x^2)
cosy * dy/dx = e^(x^2) * 2x
dydx = 2xe^(x^2)/cosy

7)
∫ cos^3(x) * sin^25(x) dx
= ∫ sin^25(x) * cos^2(x) * cosx dx

Let u = sinx.
dx = du/cosx

= ∫ u^25 * (1 - u^2) du
= ∫ u^25 - u^27 du
= 1/26 u^26 - 1/28 u^28 + C
= 1/26 sin^26(x) - 1/28 sin^28(x) + C

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