Câu hỏi của Nguyễn Trần Khánh Linh

Question

tính nguyên hàm

1.$\int\dfrac{\cos x}{3\sin x-7}dx$

2. $\int\sin x.$e^(2$\cos x$+3)dx

3. $\int\dfrac{\sin x+x\cos x}{\left(x\sin x\right)^2}dx$

(bằng pp đổi biến)

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Akai Haruma · Accepted Answer

1)

Ta có $P_1=\int \frac{\cos xdx}{2\sin x-7}=\int \frac{d(\sin x)}{3\sin x-7}$

Đặt $\sin x=t\Rightarrow P_1=\int \frac{dt}{3t-7}=\frac{1}{3}\int \frac{d(3t-7)}{3t-7}=\frac{1}{3}\ln |3t-7|+c$

$=\frac{1}{3}\ln |3\sin x-7|+c$

2)

$P_2=\int \sin xe^{2\cos x+3}dx$

Đặt $\cos x=t$

$P_2=-\int e^{2\cos x+3}d(\cos x)=-\int e^{2t+3}dt$

$=-\frac{1}{2}\int e^{2t+3}d(2t+3)=\frac{-1}{2}e^{2t+3}+c$

$=\frac{-e^{2\cos x+3}}{2}+c$Câu trả lời: 1)

Ta có $P_1=\int \frac{\cos xdx}{2\sin x-7}=\int \frac{d(\sin x)}{3\sin x-7}$

Đặt $\sin x=t\Rightarrow P_1=\int \frac{dt}{3t-7}=\frac{1}{3}\int \frac{d(3t-7)}{3t-7}=\frac{1}{3}\ln |3t-7|+c$

$=\frac{1}{3}\ln |3\sin x-7|+c$

2)

$P_2=\int \sin xe^{2\cos x+3}dx$

Đặt $\cos x=t$

$P_2=-\int e^{2\cos x+3}d(\cos x)=-\int e^{2t+3}dt$

$=-\frac{1}{2}\int e^{2t+3}d(2t+3)=\frac{-1}{2}e^{2t+3}+c$

$=\frac{-e^{2\cos x+3}}{2}+c$

Akai Haruma · Answer

3)

$P_3=\int \frac{\sin x+x\cos x}{(x\sin x)^2}dx$

Để ý rằng $(x\sin x)'=x'\sin x+x(\sin x)'=\sin x+x\cos x$

Do đó: $d(x\sin x)=(x\sin x)'dx=(\sin x+x\cos x)dx$

Suy ra $P_3=\int \frac{d(x\sin x)}{(x\sin x)^2}$

Đặt $x\sin x=t\Rightarrow P_3=\int \frac{dt}{t^2}=\frac{-1}{t}+c=\frac{-1}{x\sin x}+c$Câu trả lời: 3)

$P_3=\int \frac{\sin x+x\cos x}{(x\sin x)^2}dx$

Để ý rằng $(x\sin x)'=x'\sin x+x(\sin x)'=\sin x+x\cos x$

Do đó: $d(x\sin x)=(x\sin x)'dx=(\sin x+x\cos x)dx$

Suy ra $P_3=\int \frac{d(x\sin x)}{(x\sin x)^2}$

Đặt $x\sin x=t\Rightarrow P_3=\int \frac{dt}{t^2}=\frac{-1}{t}+c=\frac{-1}{x\sin x}+c$

Bài 1: Nguyên hàm