Tính nguyên hàm của các hàm sau:
1. \(\int sin^2\)\(\dfrac{x}{2}\) dx
2. \(\int cos^23x\) dx
3. \(\int4cos^2\dfrac{x}{2}\) dx
Tính nguyên hàm các hàm số sau:
1. \(I=\int\dfrac{cos^2x}{sin^8x}dx\)
2. \(I=\int\left(e^{sinx}+cosx\right)cosxdx\)
1.
\(I=\int\dfrac{cot^2x}{sin^6x}dx=\int\dfrac{cot^2x}{sin^4x}.\dfrac{1}{sin^2x}=\int cot^2x\left(1+cot^2x\right)^2.\dfrac{1}{sin^2x}dx\)
Đặt \(u=cotx\Rightarrow du=-\dfrac{1}{sin^2x}dx\)
\(I=-\int u^2\left(1+u^2\right)^2du=-\int\left(u^6+2u^4+u^2\right)du\)
\(=-\dfrac{1}{7}u^7+\dfrac{2}{5}u^5+\dfrac{1}{3}u^3+C\)
\(=-\dfrac{1}{7}cot^7x+\dfrac{2}{5}cot^5x+\dfrac{1}{3}cot^3x+C\)
2.
\(I=\int\left(e^{sinx}+cosx\right).cosxdx=\int e^{sinx}.cosxdx+\int cos^2xdx\)
\(=\int e^{sinx}.d\left(sinx\right)+\dfrac{1}{2}\int\left(1+cos2x\right)dx\)
\(=e^{sinx}+\dfrac{1}{2}x+\dfrac{1}{4}sin2x+C\)
Tìm nguyên hàm của các hàm số sau:
a) \(\int\left(6x-\dfrac{1}{sin^2x}+1\right)dx\)
b) \(\int\dfrac{x^3+2x^2-1}{x^2}dx\)
Tính các nguyên hàm sau :
a) \(\int x\left(3-x\right)^5dx\)
b) \(\int\left(2^x-3^x\right)^2dx\)
c) \(\int x\sqrt{2-5x}dx\)
d) \(\int\dfrac{\ln\left(\cos x\right)}{\cos^2x}dx\)
e) \(\int\dfrac{x}{\sin^2x}dx\)
\(\int\dfrac{x+1}{\left(x-2\right)\left(x+3\right)}dx\)
h) \(\int\dfrac{1}{1-\sqrt{x}}dx\)
i) \(\int\sin3x\cos2xdx\)
k) \(\int\dfrac{\sin^3x}{\cos^2x}dx\)
l) \(\int\dfrac{\sin x\cos x}{\sqrt{a^2\sin^2x+b^2\cos^2x}}dx\) (\(a^2\ne b^2\))
Tính nguyên hàm của:
1, \(\int\)\(\dfrac{x^3}{x-2}dx\)
2, \(\int\)\(\dfrac{dx}{x\sqrt{x^2+1}}\)
3, \(\int\)\((\dfrac{5}{x}+\sqrt{x^3})dx\)
4, \(\int\)\(\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx\)
5, \(\int\)\(\dfrac{dx}{\sqrt{1-x^2}}\)
a. \(\int\dfrac{x^3}{x-2}dx=\int\left(x^2+2x+4+\dfrac{8}{x-2}\right)dx=\dfrac{1}{3}x^3+x^2+4x+8ln\left|x-2\right|+C\)
b. \(\int\dfrac{dx}{x\sqrt{x^2+1}}=\int\dfrac{xdx}{x^2\sqrt{x^2+1}}\)
Đặt \(\sqrt{x^2+1}=u\Rightarrow x^2=u^2-1\Rightarrow xdx=udu\)
\(I=\int\dfrac{udu}{\left(u^2-1\right)u}=\int\dfrac{du}{u^2-1}=\dfrac{1}{2}\int\left(\dfrac{1}{u-1}-\dfrac{1}{u+1}\right)du=\dfrac{1}{2}ln\left|\dfrac{u-1}{u+1}\right|+C\)
\(=\dfrac{1}{2}ln\left|\dfrac{\sqrt{x^2+1}-1}{\sqrt{x^2+1}+1}\right|+C\)
c. \(\int\left(\dfrac{5}{x}+\sqrt{x^3}\right)dx=\int\left(\dfrac{5}{x}+x^{\dfrac{3}{2}}\right)dx=5ln\left|x\right|+\dfrac{2}{5}\sqrt{x^5}+C\)
d. \(\int\dfrac{x\sqrt{x}+\sqrt{x}}{x^2}dx=\int\left(x^{-\dfrac{1}{2}}+x^{-\dfrac{3}{2}}\right)dx=2\sqrt{x}-\dfrac{1}{2\sqrt{x}}+C\)
e. \(\int\dfrac{dx}{\sqrt{1-x^2}}=arcsin\left(x\right)+C\)
Bằng cách biến đổi các hàm số lượng giác, hãy tính :
a) \(\int\sin^4xdx\)
b) \(\int\dfrac{1}{\sin^3x}dx\)
c) \(\int\sin^3x\cos^4xdx\)
d) \(\int\sin^4x\cos^4xdx\)
e) \(\int\dfrac{1}{\cos x\sin^2x}dx\)
g) \(\int\dfrac{1+\sin x}{1+\cos x}dx\)
a) \(\sin^4x=\left(\sin^2x\right)^2=\left(\dfrac{1-\cos2x}{2}\right)^2\)
\(=\dfrac{1}{4}\left(1-2\cos2x+\cos^22x\right)\)
\(=\dfrac{1}{4}\left(1-2.\cos2x+\dfrac{1+\cos4x}{2}\right)\)
\(=\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\)
Vậy:
\(\int\sin^4x\text{dx}=\int\left(\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\right)\text{dx}\)
\(=\dfrac{3}{8}x-\dfrac{1}{4}\sin2x+\dfrac{1}{32}\sin4x+C\)
Tính các nguyên hàm sau bằng phương pháp đổi biến số :
a) \(\int x^2\sqrt[3]{1+x^3}dx\) với \(x>-1\) (đặt \(t=1+x^3\))
b) \(\int xe^{-x^2}dx\) (đặt \(t=x^2\))
c) \(\int\dfrac{x}{\left(1+x^2\right)^2}dx\) (đặt \(t=1+x^2\))
d) \(\int\dfrac{1}{\left(1-x\right)\sqrt{x}}dx\) (đặt \(t=\sqrt{x}\))
e) \(\int\sin\dfrac{1}{x}.\dfrac{1}{x^2}dx\) (đặt \(t=\dfrac{1}{x}\))
g) \(\int\dfrac{\left(\ln x\right)^2}{x}dx\) (đặt \(t=\ln x\))
h) \(\int\dfrac{\sin x}{\sqrt[3]{\cos^2x}}dx\) (đặt \(t=\cos x\) )
i) \(\int\cos x\sin^3xdx\) (đặt \(t=\sin x\))
k) \(\int\dfrac{1}{e^x-e^{-x}}dx\) (đặt \(t=e^x\) )
l) \(\int\dfrac{\cos x+\sin x}{\sqrt{\sin x-\cos x}}dx\) (đặt \(t=\sin x-\cos x\))
Tính các tích phân sau :
a) \(\int\limits^{\dfrac{\pi}{4}}_0\cos2x.\cos^2xdx\)
b) \(\int\limits^1_{\dfrac{1}{2}}\dfrac{e^x}{e^{2x}-1}dx\)
c) \(\int\limits^1_0\dfrac{x+2}{x^2+2x+1}\ln\left(x+1\right)dx\)
d) \(\int\limits^{\dfrac{\pi}{4}}_0\dfrac{x\sin x+\left(x+1\right)\cos x}{x\sin x+\cos x}dx\)
a)
Ta có \(A=\int ^{\frac{\pi}{4}}_{0}\cos 2x\cos^2xdx=\frac{1}{4}\int ^{\frac{\pi}{4}}_{0}\cos 2x(\cos 2x+1)d(2x)\)
\(\Leftrightarrow A=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos x(\cos x+1)dx=\frac{1}{4}\int ^{\frac{\pi}{2}}_{0}\cos xdx+\frac{1}{8}\int ^{\frac{\pi}{2}}_{0}(\cos 2x+1)dx\)
\(\Leftrightarrow A=\frac{1}{4}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin x+\frac{1}{16}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|\sin 2x+\frac{1}{8}\left.\begin{matrix} \frac{\pi}{2}\\ 0\end{matrix}\right|x=\frac{1}{4}+\frac{\pi}{16}\)
b)
\(B=\int ^{1}_{\frac{1}{2}}\frac{e^x}{e^{2x}-1}dx=\frac{1}{2}\int ^{1}_{\frac{1}{2}}\left ( \frac{1}{e^x-1}-\frac{1}{e^x+1} \right )d(e^x)\)
\(\Leftrightarrow B=\frac{1}{2}\left.\begin{matrix} 1\\ \frac{1}{2}\end{matrix}\right|\left | \frac{e^x-1}{e^x+1} \right |\approx 0.317\)
c)
Có \(C=\int ^{1}_{0}\frac{(x+2)\ln(x+1)}{(x+1)^2}d(x+1)\).
Đặt \(x+1=t\)
\(\Rightarrow C=\int ^{2}_{1}\frac{(t+1)\ln t}{t^2}dt=\int ^{2}_{1}\frac{\ln t}{t}dt+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
\(=\int ^{2}_{1}\ln td(\ln t)+\int ^{2}_{1}\frac{\ln t}{t^2}dt=\frac{\ln ^22}{2}+\int ^{2}_{1}\frac{\ln t}{t^2}dt\)
Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=\frac{dt}{t^2}\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=\frac{-1}{t}\end{matrix}\right.\Rightarrow \int ^{2}_{1}\frac{\ln t}{t^2}dt=\left.\begin{matrix} 2\\ 1\end{matrix}\right|-\frac{\ln t+1}{t}=\frac{1}{2}-\frac{\ln 2 }{2}\)
\(\Rightarrow C=\frac{1}{2}-\frac{\ln 2}{2}+\frac{\ln ^22}{2}\)
d)
\(D=\int ^{\frac{\pi}{4}}_{0}\frac{x\sin x+(x+1)\cos x}{x\sin x+\cos x}dx=\int ^{\frac{\pi}{4}}_{0}dx+\int ^{\frac{\pi}{4}}_{0}\frac{x\cos x}{x\sin x+\cos x}dx\)
Ta có:
\(\int ^{\frac{\pi}{4}}_{0}dx=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|x=\frac{\pi}{4}\)
\(\int ^{\frac{\pi}{4}}_{0}\frac{x\cos xdx}{x\sin x+\cos x}=\int ^{\frac{\pi}{4}}_{0}\frac{d(x\sin x+\cos x)}{x\sin x+\cos x}=\left.\begin{matrix} \frac{\pi}{4}\\ 0\end{matrix}\right|\ln |x\sin x+\cos x|\)
\(=\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Suy ra \(D=\frac{\pi}{4}+\ln|\frac{\pi\sqrt{2}}{8}+\frac{\sqrt{2}}{2}|\)
Tính các nguyên hàm sau :
a) \(\int\left(2x-3\right)\sqrt{x-3}dx\), đặt \(u=\sqrt{x-3}\)
b) \(\int\dfrac{x}{\left(1+x^2\right)^{\dfrac{3}{2}}}dx\) , đặt \(u=\sqrt{x^2+1}\)
c) \(\int\dfrac{e^x}{e^x+e^{-x}}dx\), đặt \(u=e^{2x}+1\)
d) \(\int\dfrac{1}{\sin x-\sin a}dx\)
e) \(\int\sqrt{x}\sin\sqrt{x}dx,\) đặt \(t=\sqrt{x}\)
g) \(\int x\ln\dfrac{x}{1+x}dx\)
a)
Đặt \(u=\sqrt{x-3}\Rightarrow x=u^2+3\)
\(I_1=\int (2x-3)\sqrt{x-3}dx=\int (2u^2+3)ud(u^2+3)=2\int (2u^2+3)u^2du\)
\(\Leftrightarrow I_1=4\int u^4du+6\int u^2du=\frac{4u^5}{5}+2u^3+c\)
b)
\(I_2=\int \frac{xdx}{\sqrt{(x^2+1)^3}}=\frac{1}{2}\int \frac{d(x^2+1)}{\sqrt{(x^2+1)^2}}\)
Đặt \(u=\sqrt{x^2+1}\). Khi đó:
\(I_2=\frac{1}{2}\int \frac{d(u^2)}{u^3}=\int \frac{udu}{u^3}=\int \frac{du}{u^2}=\frac{-1}{u}+c\)
c)
\(I_3=\int \frac{e^xdx}{e^x+e^{-x}}=\int \frac{e^{2x}dx}{e^{2x}+1}=\frac{1}{2}\int\frac{d(e^{2x}+1)}{e^{2x}+1}\)
\(\Leftrightarrow I_3=\frac{1}{3}\ln |e^{2x}+1|+c=\frac{1}{2}\ln|u|+c\)
d)
\(I_4=\int \frac{dx}{\sin x-\sin a}=\int \frac{dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}\)
\(\Leftrightarrow I_4=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x+a}{2}-\frac{x-a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )\sin \left ( \frac{x-a}{2} \right )}=\frac{1}{\cos a}\int \frac{\cos \left ( \frac{x-a}{2} \right )dx}{2\sin \left ( \frac{x-a}{2} \right )}+\frac{1}{\cos a}\int \frac{\sin \left ( \frac{x+a}{2} \right )dx}{2\cos \left ( \frac{x+a}{2} \right )}\)
\(\Leftrightarrow I_4=\frac{1}{\cos a}\left ( \ln |\sin \frac{x-a}{2}|-\ln |\cos \frac{x+a}{2}| \right )+c\)
e)
Đặt \(t=\sqrt{x}\Rightarrow x=t^2\)
\(I_5=\int t\sin td(t^2)=2\int t^2\sin tdt\)
Đặt \(\left\{\begin{matrix} u=t^2\\ dv=\sin tdt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=2tdt\\ v=-\cos t\end{matrix}\right.\)
\(\Rightarrow I_5=-2t^2\cos t+4\int t\cos tdt\)
Tiếp tục nguyên hàm từng phần \(\Rightarrow \int t\cos tdt=t\sin t+\cos t+c\)
\(\Rightarrow I_5=-2t^2\cos t+4t\sin t+4\cos t+c\)
g)
Có \(I_6=\int x\ln \left ( \frac{x}{x+1} \right )dx=\int x\ln xdx-\int x\ln (x+1)dx\)
Đặt \(\left\{\begin{matrix} u=\ln x\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dx}{x}\\ v=\frac{x^2}{2}\end{matrix}\right.\Rightarrow \int x\ln xdx=\frac{x^2\ln x}{2}-\int \frac{xdx}{2}\)
\(\Leftrightarrow \int x\ln xdx=\frac{x^2\ln x}{2}-\frac{x^2}{4}+c\)
Tương tự, \(\int x\ln (x+1)dx=\frac{x^2\ln (x+1)}{2}-\int \frac{x^2}{2(x+1)}dx\)
\(=\frac{x^2\ln (x+1)}{2}-\frac{x^2}{4}+\frac{x}{2}-\frac{\ln (x+1)}{2}+c\)
Suy ra \(I_5=\frac{x^2}{2}\ln \frac{x}{x+1}+\frac{1}{2}\ln|x+1|-\frac{x}{2}+c\)
\(\int tan\left(x\right)-ln^{15}\left(cos\left(x\right)\right)dx\)
\(\int\dfrac{x^4+x^2+1}{2x^3+5x^2-7}dx\)
tính nguyên hàm , ai giúp mình 2 bài này với hoặc 1 bài thôi cũng đc ạ , xin cảm ơn nhiều.