Lời giải

Đặt \( t = \frac{\pi}{2} - x  \Rightarrow x = \frac{\pi}{2} - t \Rightarrow dx = -dt \)

\( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^5 x}{\sin^5 x + \cos^5 x} \, dx = \int_{\frac{\pi}{2}}^{0} \frac{\sin^5 \left( \frac{\pi}{2} - t \right)}{\sin^5 \left( \frac{\pi}{2} - t \right) + \cos^5 \left( \frac{\pi}{2} - t \right)} (-dt)\)

 \( = \int_{0}^{\frac{\pi}{2}} \frac{\cos^5 t}{\cos^5 t + \sin^5 t} \, dt = J \)

•   \( I + J = \int_{0}^{\frac{\pi}{2}} \, dx = \frac{\pi}{2} \Rightarrow I = \frac{\pi}{4} \)

\( \Rightarrow \begin{cases}
n = 4\\
m = 0
\end{cases}
\Rightarrow  n + m = 4 \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)

\( I = \int_{0}^{\frac{\pi}{2}} \frac{\sin^n x}{\sin^n x + \cos^n x} \, dx =  \int_{0}^{\frac{\pi}{2}} \frac{\cos^n x}{\cos^n x + \sin^n x} \, dt = \frac{\pi}{4} \)

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Lời giải

Đặt \( t = \frac{\pi}{2} - x \Rightarrow dt = -dx \)

\( \begin{cases}
x = 0 \Rightarrow t = \frac{\pi}{2} \\
x = \frac{\pi}{2} \Rightarrow t = 0
\end{cases} \)

\( I = \int_{\frac{\pi}{2}}^{0} \frac{1 + \sin^n \left(\frac{\pi}{2} - t\right)}{2 + \sin^m \left(\frac{\pi}{2} - t\right)} (-dt) = \int_{0}^{\frac{\pi}{2}} \frac{1 + \cos^n t}{2 + \cos^m t} \, dt = a \)

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{B}} \)

\( \int_{0}^{\frac{\pi}{2}} f(\sin x) \, dx = \int_{0}^{\frac{\pi}{2}} f(\cos x) \, dx \)

page 33

Lời giải

a) Vì \( \int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx \)

Xét \( \int_{0}^{a} f(x) \, dx \).

Đặt \( t = -x \implies x = -t \implies dx = -dt \)

\( \begin{cases}
x =0\\
x = a 
\end{cases}
\Rightarrow \begin{cases}
t = 0 \\
t = -a
\end{cases} \)

\( \int_{0}^{a} f(x) \, dx = \int_{0}^{-a} f(-t)(-dt) = -\int_{-a}^{0} f(-t) \, dt = -\int_{-a}^{0} f(x) \, dx \)

\( \Rightarrow \int_{-a}^{a} f(x) \, dx = 0 \)

b) Tương tự:

Ví dụ:
a) \( \int_{-\pi}^{\pi} \sin^3 x \, dx = 0 \) 

b) \( \int_{-1}^{1} (2x^3 - 4x) \, dx = 0\)

c) \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x \ cos x \, dx = 0 \)  

d) \( \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^2 \sin x \, dx = 0 \) 

page 34

Lời giải

\( \int_{-2}^{2} \left( 3x^{2017} - 4x^3 + x^2 - 2x \right) \, dx\)

\(= \int_{-2}^{2} \left( 3x- 4x^3 - 2x \right) \, dx + \int_{-2}^{2} x^2 \, dx \)

\(=  0 + \frac{x^3}{3} \bigg|_{-2}^{2} = \frac{8}{3} + \frac{8}{3} = \frac{16}{3} \)

\( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos x \, dx = \sin x \bigg|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2 \)

page 35

Lời giải

* Bấm \( \int_{-1}^3 (x^3 - 3x^2 + 2)^{2015} \, dx \) = ( Math error! )

\( I = \int_{-1}^3 (x-1)(x^2 - 2x - 1) \, dx = \int_{-1}^3 (x-1)\left[(x-1)^2 - 3\right] \, dx \)

\( = \int_{-1}^3 \left[ (x-1)^3 - 3(x-1) \right] \, dx \)

Đặt \( t = x - 1 \implies dt = dx \)  

\( \begin{cases}
x = -1 \Rightarrow  t = -2\\
x = 3  \Rightarrow  t = 2
\end{cases}\)

\( I = \int_{-2}^{2} \left( t^3 - 3t  \right) dt = 0\)

Vì \( f(t) = t^3 - 3t\) là hàm số lẻ

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