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a) \(\int\left(x+\ln x\right)x^2\text{d}x=\int x^3\text{d}x+\int x^2\ln x\text{dx}\)
\(=\dfrac{x^4}{4}+\int x^2\ln x\text{dx}+C\) (*)
Để tính: \(\int x^2\ln x\text{dx}\) ta sử dụng công thức tính tích phân từng phần như sau:
Đặt \(\left\{{}\begin{matrix}u=\ln x\\v'=x^2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=\dfrac{1}{x}\\v=\dfrac{1}{3}x^3\end{matrix}\right.\)
Suy ra:
\(\int x^2\ln x\text{dx}=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}\int x^2\text{dx}\)
\(=\dfrac{1}{3}x^3\ln x-\dfrac{1}{3}.\dfrac{1}{3}x^3\)
Thay vào (*) ta tính được nguyên hàm của hàm số đã cho bằng:
(*) \(=\dfrac{1}{3}x^3-\dfrac{1}{3}x^3\ln x+\dfrac{1}{9}x^3+C\)
\(=\dfrac{4}{9}x^3-\dfrac{1}{3}x^3\ln x+C\)
b) Đặt \(\left\{{}\begin{matrix}u=x+\sin^2x\\v'=\sin x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u'=1+2\sin x.\cos x\\v=-\cos x\end{matrix}\right.\)
Ta có:
\(\int\left(x+\sin^2x\right)\sin x\text{dx}=-\left(x+\sin^2x\right)\cos x+\int\left(1+2\sin x\cos^2x\right)\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)
\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)
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.
\(I=\int\left(e^{sinx}+cosx\right).cosxdx=\int e^{sinx}.cosxdx+\int cos^2xdx\)
\(=\int e^{sinx}.d\left(sinx\right)+\dfrac{1}{2}\int\left(1+cos2x\right)dx\)
\(=e^{sinx}+\dfrac{1}{2}x+\dfrac{1}{4}sin2x+C\)
\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)
a)
Ta có:
∫π20cos2xsin2xdx=12∫π20cos2x(1−cos2x)dx=12∫π20[cos2x−1+cos4x2]dx=14∫π20(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]π20=−14.π2=−π8∫0π2cos2xsin2xdx=12∫0π2cos2x(1−cos2x)dx=12∫0π2[cos2x−1+cos4x2]dx=14∫0π2(2cos2x−cos4x−1)dx=14[sin2x−sin4x4−x]0π2=−14.π2=−π8
b)
Ta có: Xét 2x – 2-x ≥ 0 ⇔ x ≥ 0.
Ta tách thành tổng của hai tích phân:
∫1−1|2x−2−x|dx=−∫0−1(2x−2−x)dx+∫10(2x−2−x)dx=−(2xln2+2−xln2)∣∣0−1+(2xln2+2−xln2)∣∣10=1ln2∫−11|2x−2−x|dx=−∫−10(2x−2−x)dx+∫01(2x−2−x)dx=−(2xln2+2−xln2)|−10+(2xln2+2−xln2)|01=1ln2
c)
∫21(x+1)(x+2)(x+3)x2dx=∫21x3+6x2+11x+6x2dx=∫21(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]∣∣21=(2+12+11ln2−3)−(12+6−6)=212+11ln2∫12(x+1)(x+2)(x+3)x2dx=∫12x3+6x2+11x+6x2dx=∫12(x+6+11x+6x2)dx=[x22+6x+11ln|x|−6x]|12=(2+12+11ln2−3)−(12+6−6)=212+11ln2
d)
∫201x2−2x−3dx=∫201(x+1)(x−3)dx=14∫20(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]∣∣20=14[1−ln2−ln3]=14(1−ln6)∫021x2−2x−3dx=∫021(x+1)(x−3)dx=14∫02(1x−3−1x+1)dx=14[ln|x−3|−ln|x+1|]|02=14[1−ln2−ln3]=14(1−ln6)
e)
∫π20(sinx+cosx)2dx=∫π20(1+sin2x)dx=[x−cos2x2]∣∣π20=π2+1∫0π2(sinx+cosx)2dx=∫0π2(1+sin2x)dx=[x−cos2x2]|0π2=π2+1
g)
I=∫π0(x+sinx)2dx∫π0(x2+2xsinx+sin2x)dx=[x33]∣∣π0+2∫π0xsinxdx+12∫π0(1−cos2x)dxI=∫0π(x+sinx)2dx∫0π(x2+2xsinx+sin2x)dx=[x33]|0π+2∫0πxsinxdx+12∫0π(1−cos2x)dx
Tính :J=∫π0xsinxdxJ=∫0πxsinxdx
Đặt u = x ⇒ u’ = 1 và v’ = sinx ⇒ v = -cos x
Suy ra:
J=[−xcosx]∣∣π0+∫π0cosxdx=π+[sinx]∣∣π0=πJ=[−xcosx]|0π+∫0πcosxdx=π+[sinx]|0π=π
Do đó:
I=π33+2π+12[x−sin2x2]∣∣π30=π33+2π+π2=2π3+15π6
a/ Tích phân này làm sao giải được nhỉ?
b/ Đặt \(\sqrt{x}=t\Rightarrow x=t^2\Rightarrow dx=2t.dt\)
\(I=\int\frac{2t^2.dt}{4-t^4}=\int\left(\frac{1}{2-t^2}-\frac{1}{2+t^2}\right)dt=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{t}+C\)
\(=\frac{1}{2\sqrt{2}}ln\left|\frac{\sqrt{2}+\sqrt{x}}{\sqrt{2}-\sqrt{x}}\right|+\frac{1}{\sqrt{2}}arctan\frac{\sqrt{2}}{\sqrt{x}}+C\)
c/ \(I=\int\frac{\sqrt{1+x^2}}{x^2}.xdx\)
Đặt \(\sqrt{1+x^2}=t\Rightarrow x^2=t^2-1\Rightarrow xdx=tdt\)
\(\Rightarrow I=\int\frac{t^2dt}{t^2-1}=\int\left(1+\frac{1}{t^2-1}\right)dt=t+ln\left|\frac{t-1}{t+1}\right|+C=\sqrt{1+x^2}+ln\left|\frac{\sqrt{1+x^2}-1}{\sqrt{1+x^2}+1}\right|+C\)
d/ Con nguyên hàm này cũng không tính được, chắc bạn ghi nhầm đề
Lời giải:
Câu 1:
\(A=\int\frac{dx}{1+\sin x}=\int \frac{(1-\sin x)dx}{1-\sin^2 x}=\int\frac{(1-\sin x)dx}{\cos ^2x}=\int\frac{dx}{\cos ^2x}-\int\frac{\sin x dx}{\cos^2 x}\)
\(\Leftrightarrow A=\int d(\tan x)+\int\frac{d(\cos x)}{\cos^2 x}=\tan x-\frac{1}{\cos x}+c\)
Câu 2:
\(B=\int \sin ^4 xdx=\int \sin^2 x(1-\cos ^2x)dx=\int \sin^2 xdx-\int \sin^2 x\cos^2xdx\)
Ta thấy \(\int \sin^2xdx=\frac{1}{2}\int (1-\cos 2x)dx=\frac{x}{2}-\frac{\sin 2x}{4}+c\)
Và \(\int \sin ^2x\cos^2xdx=\frac{1}{4}\int \sin^22xdx=\frac{1}{8}\int (1-\cos4x)dx=\frac{x}{8}-\frac{\sin 4x}{32}+c\)
\(\Rightarrow B=\frac{3}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+c\)
Câu 3:
\(C=\int (\sin ^6 x+\cos^6 x)dx=\int (\sin^2x+\cos^2x)[\sin^4x-\sin^2x\cos^2x+\cos^4x)dx\)
\(\Leftrightarrow C=\int [(\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x]dx\)
\(\Leftrightarrow C=x-\frac{3x}{8}+\frac{3\sin 4x}{32}+c=\frac{5x}{8}+\frac{3\sin 4x}{32}+c\)