Tính \(I=\int\tan^4xdx\)
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GV
Giáo viên Toán
Giáo viên
4 tháng 5 2017
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:
\(\int\sin^4x\text{dx}=\int\left(\dfrac{3}{8}-\dfrac{1}{2}\cos2x+\dfrac{1}{8}\cos4x\right)\text{dx}\)
\(=\dfrac{3}{8}x-\dfrac{1}{4}\sin2x+\dfrac{1}{32}\sin4x+C\)
AH
Akai Haruma
Giáo viên
6 tháng 3 2017
Câu 1)
Ta có \(I=\int ^{1}_{0}\frac{dx}{\sqrt{3+2x-x^2}}=\int ^{1}_{0}\frac{dx}{4-(x-1)^2}\).
Đặt \(x-1=2\cos t\Rightarrow \sqrt{4-(x-1)^2}=\sqrt{4-4\cos^2t}=2|\sin t|\)
Khi đó:
\(I=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{d(2\cos t+1)}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}\frac{2\sin tdt}{2\sin t}=\int ^{\frac{2\pi}{3}}_{\frac{\pi}{2}}dt=\left.\begin{matrix} \frac{2\pi}{3}\\ \frac{\pi}{2}\end{matrix}\right|t=\frac{\pi}{6}\)
Câu 3)
\(K=\int ^{3}_{2}\ln (x^3-3x+2)dx=\int ^{3}_{2}\ln [(x+2)(x-1)^2]dx\)
\(=\int ^{3}_{2}\ln (x+2)d(x+2)+2\int ^{3}_{2}\ln (x-1)d(x-1)\)
Xét \(\int \ln tdt\): Đặt \(\left\{\begin{matrix} u=\ln t\\ dv=dt\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{dt}{t}\\ v=t\end{matrix}\right.\Rightarrow \int \ln t dt=t\ln t-t\)
\(\Rightarrow K=\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x+2)[\ln (x+2)-1]+2\left.\begin{matrix} 3\\ 2\end{matrix}\right|(x-1)[\ln (x-1)-1]\)
\(=5\ln 5-4\ln 4-1+4\ln 2-2=5\ln 5-4\ln 2-3\)
AH
Akai Haruma
Giáo viên
6 tháng 3 2017
Bài 2)
\(J=\int ^{1}_{0}x\ln (2x+1)dx\). Đặt \(\left\{\begin{matrix} u=\ln (2x+1)\\ dv=xdx\end{matrix}\right.\Rightarrow \left\{\begin{matrix} du=\frac{2dx}{2x+1}\\ v=\frac{x^2}{2}\end{matrix}\right.\)
Khi đó:
\(J=\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2\ln (2x+1)}{2}-\int ^{1}_{0}\frac{x^2}{2x+1}dx\)\(=\frac{\ln 3}{2}-\frac{1}{4}\int ^{1}_{0}(2x-1+\frac{1}{2x+1})dx\)
\(=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{x^2-x}{4}-\frac{1}{8}\int ^{1}_{0}\frac{d(2x+1)}{2x+1}=\frac{\ln 3}{2}-\left.\begin{matrix} 1\\ 0\end{matrix}\right|\frac{\ln (2x+1)}{8}\)
\(=\frac{\ln 3}{2}-\frac{\ln 3}{8}=\frac{3\ln 3}{8}\)
CM
13 tháng 11 2019
Chọn C
Đặt u = 2 x + 3 d v = sin 4 x . d x ⇒ d u = 2 . d x v = - 1 4 cos 4 x . d x
⇒ I = - 1 4 ( 2 x + 3 ) cos 4 x | 0 π 4 + 1 2 ∫ 0 π 4 cos 4 x d x = - 1 4 ( 2 x + 3 ) cos 4 x + 1 2 . 1 4 . sin 4 x | 0 π 4 = π 8 + 3 2
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 1 2022
\(\int tan^3xdx=\int tan^2x.tanxdx=\int\left(\dfrac{1}{cos^2x}-1\right)tanxdx\)
\(=\int tanx.d\left(tanx\right)-\int tanxdx=\dfrac{1}{2}tan^2x-ln\left|cosx\right|+C\)
5 tháng 4 2016
Ta có \(I=\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{\ln2.\ln\left(2\tan x\right)}{\sin2x.\ln\left(2\tan x\right)}dx=\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}+\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}\)
Tính \(\ln2\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x.\ln\left(2\tan x\right)}=\frac{\ln2}{2}\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{d\left[\ln\left(2\tan x\right)\right]}{\ln2\left(2\tan x\right)}=\frac{\ln2}{2}\left[\ln\left(\ln\left(2\tan x\right)\right)\right]|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{\ln2}{2}.\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)\)
Tính \(\int\limits^{\frac{\pi}{3}}_{\frac{\pi}{4}}\frac{dx}{\sin2x}=\frac{1}{2}\ln\left(\tan x\right)|^{\frac{\pi}{3}}_{\frac{\pi}{4}}=\frac{1}{2}\ln\sqrt{3}\)
Vậy \(I=\frac{\ln2}{2}\ln\left(\frac{\ln2\sqrt{3}}{\ln2}\right)+\frac{1}{2}\ln\sqrt{3}\)
PT
1
TA
1
23 tháng 1 2016
Ta biến đổi f(x) về dạng :
\(f\left(x\right)=\frac{\sin x.\sin\left(x+\frac{\pi}{4}\right)+\cos x.\cos\left(x+\frac{\pi}{4}\right)}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}-1=\frac{\cos\frac{\pi}{4}}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}-1\)
\(\Rightarrow F\left(x\right)=\frac{\sqrt{2}}{2}\int\frac{dx}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}dx-\int dx=\frac{\sqrt{2}}{2}\int\frac{dx}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}dx-x\left(1\right)\)
Để tính \(J=\int\frac{dx}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}dx\)
Ta có \(\int\frac{dx}{\cos x.\cos\left(x+\frac{\pi}{4}\right)}dx=\sqrt{2}\int\frac{1}{\cos x.\left(\cos x-\sin x\right)}dx=\sqrt{2}\int\frac{1}{\left(1-\tan x\right)}.\frac{1}{\cos^2x}dx\)
\(=-\sqrt{2}\int\frac{d\left(1-\tan x\right)}{1-\tan x}=\sqrt{2}\ln\left|1-\tan x\right|+C\)
\(I=\int tan^2x\left(\dfrac{1}{cos^2x}-1\right)dx=\int tan^2x.\dfrac{1}{cos^2x}dx-\int tan^2xdx\)
\(=\int tan^2x.d\left(tanx\right)-\int\left(\dfrac{1}{cos^2x}-1\right)dx=\dfrac{1}{3}tan^3x-tanx+x+C\)