Chủ đề này có 1 trả lời

[queo]

1. I= $\int\limits_{0}^{1} \dfrac{x^2e^x}{x^2+4x+4}dx$
2. J= $\int\limits_{0}^{ \dfrac{ \pi }{4} } \dfrac{4cosx+3sinx}{cosx+2sinx} dx$
3.L= $\int\limits_{0}^{ \dfrac{ \pi }{4} }ln(1+tanx)dx$

[stuart clark]

(3) $I = \int_{0}^{\dfrac{\pi}{4}}ln(1+tan\theta)d\theta......................................(1)$

$I = \int_{0}^{\dfrac{\pi}{4}}ln\left(1+tan(\dfrac{\pi}{4}-\theta)\right)d\theta............................(2)$

Here We have use the formula $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a-x)dx$

Now Add $(1)$ and $(2)$, We get

$2I = \int_{0}^{\dfrac{\pi}{4}}ln(1+tan\theta)d\theta+ln\left(1+tan(\dfrac{\pi}{4}-\theta)\right)d\theta$

Now Let $\theta =A$ and $(\dfrac{\pi}{4}-\theta) =B$. Then $(1+tan\; A).(1+tan\;B) = 2$ because $A+B=\dfrac{\pi}{4}$

$2I = \int_{0}^{\dfrac{\pi}{4}}ln\left((1+tan\;A).(1+tan\; B)\right)d\theta $

$2I = \int_{0}^{\dfrac{\pi}{4}}ln(2)d\theta = \dfrac{\pi}{4}.ln(2)$

$\boxed{I = \dfrac{\pi}{8}ln(2)}$