Giup minh voi a $a,~\int^B_0(4\sin x+3\cos x)dx$,b, Stanx dx,c, ( P2 cot'x dx,7% d, S e'dx

Question

Giup minh voi a $a,~\int^B_0(4\sin x+3\cos x)dx$,b, Stanx dx,c, ( P2 cot&#x27;x dx,7% d, S e&#x27;dx

Accepted Answer

a) Tính $\int ^{\frac{\pi}{4}}_0 (4\sin x + 3 \cos x) \, dx$

Ta có:
$$
\int (4\sin x + 3 \cos x) \, dx = -4\cos x + 3\sin x + C
$$

Do đó:
$$
\int ^{\frac{\pi}{4}}_0 (4\sin x + 3 \cos x) \, dx = \left[ -4\cos x + 3\sin x \right] ^{\frac{\pi}{4}}_0
$$
$$
= \left( -4\cos \frac{\pi}{4} + 3\sin \frac{\pi}{4} \right) - \left( -4\cos 0 + 3\sin 0 \right)
$$
$$
= \left( -4 \cdot \frac{\sqrt{2}}{2} + 3 \cdot \frac{\sqrt{2}}{2} \right) - \left( -4 \cdot 1 + 3 \cdot 0 \right)
$$
$$
= \left( -2\sqrt{2} + \frac{3\sqrt{2}}{2} \right) - (-4)
$$
$$
= -2\sqrt{2} + \frac{3\sqrt{2}}{2} + 4
$$
$$
= -\frac{4\sqrt{2}}{2} + \frac{3\sqrt{2}}{2} + 4
$$
$$
= -\frac{\sqrt{2}}{2} + 4
$$
$$
= 4 - \frac{\sqrt{2}}{2}
$$

b) Tính $\int ^{\frac{\pi}{4}}_0 \tan^3 x \, dx$

Ta có:
$$
\tan^3 x = \tan x \cdot \tan^2 x = \tan x (\sec^2 x - 1)
$$
$$
\int \tan^3 x \, dx = \int \tan x (\sec^2 x - 1) \, dx = \int \tan x \sec^2 x \, dx - \int \tan x \, dx
$$

Xét từng phần:
$$
\int \tan x \sec^2 x \, dx = \int u \, du = \frac{u^2}{2} = \frac{\tan^2 x}{2}
$$
$$
\int \tan x \, dx = -\ln |\cos x|
$$

Do đó:
$$
\int \tan^3 x \, dx = \frac{\tan^2 x}{2} + \ln |\cos x| + C
$$

Vậy:
$$
\int ^{\frac{\pi}{4}}_0 \tan^3 x \, dx = \left[ \frac{\tan^2 x}{2} + \ln |\cos x| \right] ^{\frac{\pi}{4}}_0
$$
$$
= \left( \frac{\tan^2 \frac{\pi}{4}}{2} + \ln |\cos \frac{\pi}{4}| \right) - \left( \frac{\tan^2 0}{2} + \ln |\cos 0| \right)
$$
$$
= \left( \frac{1^2}{2} + \ln \frac{\sqrt{2}}{2} \right) - \left( 0 + \ln 1 \right)
$$
$$
= \frac{1}{2} + \ln \frac{\sqrt{2}}{2}
$$
$$
= \frac{1}{2} + \ln \sqrt{2} - \ln 2
$$
$$
= \frac{1}{2} + \frac{1}{2}\ln 2 - \ln 2
$$
$$
= \frac{1}{2} - \frac{1}{2}\ln 2
$$

c) Tính $\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^2 x \, dx$

Ta có:
$$
\cot^2 x = \csc^2 x - 1
$$
$$
\int \cot^2 x \, dx = \int (\csc^2 x - 1) \, dx = -\cot x - x + C
$$

Do đó:
$$
\int ^{\frac{\pi}{2}}_{\frac{\pi}{4}} \cot^2 x \, dx = \left[ -\cot x - x \right] ^{\frac{\pi}{2}}_{\frac{\pi}{4}}
$$
$$
= \left( -\cot \frac{\pi}{2} - \frac{\pi}{2} \right) - \left( -\cot \frac{\pi}{4} - \frac{\pi}{4} \right)
$$
$$
= \left( 0 - \frac{\pi}{2} \right) - \left( -1 - \frac{\pi}{4} \right)
$$
$$
= -\frac{\pi}{2} + 1 + \frac{\pi}{4}
$$
$$
= 1 - \frac{\pi}{4}
$$

d) Tính $\int ^{-1}_{-2} e^{x+2} \, dx$

Ta có:
$$
\int e^{x+2} \, dx = e^{x+2} + C
$$

Do đó:
$$
\int ^{-1}_{-2} e^{x+2} \, dx = \left[ e^{x+2} \right] ^{-1}_{-2}
$$
$$
= e^{-1+2} - e^{-2+2}
$$
$$
= e^1 - e^0
$$
$$
= e - 1
$$

Đáp số:
a) $4 - \frac{\sqrt{2}}{2}$
b) $\frac{1}{2} - \frac{1}{2}\ln 2$
c) $1 - \frac{\pi}{4}$
d) $e - 1$