TA CÓ

\(\int\left(x^3-2\right)^2xdx=\int\left(x^6-4x^3+4\right)xdx=\int\left(x^7-4x^4+4x\right)dx=\frac{x^8}{8}-\frac{4x^5}{5}+2x^2+C\)

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.

\(I=\int\left(e^{sinx}+cosx\right).cosxdx=\int e^{sinx}.cosxdx+\int cos^2xdx\)

\(=\int e^{sinx}.d\left(sinx\right)+\dfrac{1}{2}\int\left(1+cos2x\right)dx\)

\(=e^{sinx}+\dfrac{1}{2}x+\dfrac{1}{4}sin2x+C\)

\(a=\int\dfrac{1}{2tan^2x+5tanx+2}.\dfrac{dx}{cos^2x}\)

Đặt \(tanx=t\Rightarrow dt=\dfrac{dx}{cos^2x}\)

\(I=\int\dfrac{dt}{2t^2+5t+2}=\int\dfrac{dt}{\left(t+2\right)\left(2t+1\right)}=\dfrac{2}{3}\int\left(\dfrac{1}{2t+1}-\dfrac{1}{2t+4}\right)dt\)

\(=\dfrac{1}{3}ln\left|\dfrac{2t+1}{2t+4}\right|+C=\dfrac{1}{3}ln\left|\dfrac{2tanx+1}{2tanx+4}\right|+C\)

Câu b hoàn toàn tương tự

\(\int sin^2\dfrac{x}{2}dx=\int\left(\dfrac{1}{2}-\dfrac{1}{2}cosx\right)dx=\dfrac{1}{2}x-\dfrac{1}{2}sinx+C\)

\(\int cos^23xdx=\int\left(\dfrac{1}{2}+\dfrac{1}{2}cos6x\right)dx=\dfrac{1}{2}x+\dfrac{1}{12}sin6x+C\)

\(\int4cos^2\dfrac{x}{2}dx=\int\left(2+2cosx\right)dx=2x+2sinx+C\)

a) Áp dụng  đồng nhất thức  \(\cos^2x+\sin^2x=1\)

ta có : \(\int\frac{1}{\cos^2x.\sin^2x}dx=\int\frac{\cos^2x+\sin^2x}{\cos^2x.\sin^2x}dx=\int\frac{dx}{\sin^2x}+\int\frac{dx}{\cos^2x}\)

                                   \(=-\cot x+\tan x+C\)

b) Khai triển biểu thức dưới dấu nguyên hàm ta thu được :

\(\int\left(\tan x+\cot x\right)^2dx=\int\left(\tan^2x+2+\cot^2x\right)dx\)

                                 \(=\int\left[\left(\tan^2x+1\right)+\left(\cot^2x+1\right)\right]dx\)

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

                                 \(=\tan x-\cot x+C\)

c) \(\int\tan^2xdx=\int\left(\frac{1}{\cos^2x}-1\right)dx=\tan x-x+C\)

d) \(\int\left(5^{3x}+\frac{1}{\sin^2\left(2x+1\right)}+\frac{1}{\sqrt[5]{4x-1}}\right)dx=\)

                                                        \(=\int5^{3x}dx+\int\frac{dx}{\sin^2\left(2x+1\right)}+\int\frac{dx}{\sqrt[5]{4x-1}}\)

                                                        \(=\frac{1}{3}\int5^{3x}d\left(3x\right)+\frac{1}{2}\int\frac{d\left(2x+1\right)}{\sin^2\left(2x+1\right)}+\frac{1}{4}\int\left(4x-1\right)^{-\frac{1}{5}}d\left(4x-1\right)\)

                                                        \(=\frac{5^{3x}}{3\ln5}-\frac{1}{2}\cot\left(2x+1\right)+\frac{5}{16}\sqrt[5]{\left(4x-1\right)^4+C}\)

\(\int\limits^{\frac{\pi}{2}}_0x.\sin^2xdx=\int\limits^{\frac{\pi}{2}}_0x\left(\frac{1-\cos2x}{2}\right)dx=\frac{1}{2}\left[\int\limits^{\frac{\pi}{2}}_0xdx-\int\limits^{\frac{\pi}{2}}_0x.\cos3xdx\right]\)

                   \(=\frac{1}{2}\left(\frac{1}{2}x^2|^{\frac{\pi}{2}}_0-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_0x.d\left(\sin2x\right)\right)\)

                   \(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(x.\sin2x\right)|^{\frac{\pi}{2}}_0-\int\limits^{\frac{\pi}{2}}_0\sin2xdx\right]\)

                  \(=\frac{1}{2}\left[\frac{\pi^2}{8}-\frac{1}{2}\left(0+\frac{1}{2}\cos2x|^{\frac{\pi}{2}}_0\right)\right]=\frac{\pi^2+8}{16}\)

4 câu 1,3,4,5 giống nhau, mình làm 1 câu và bạn dựa vào đó tự xử lý mấy câu còn lại nhé

1/ \(I=\int sin2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=sin2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=2cos2x.dx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\int cos2x.e^{3x}dx=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}I_1\)

Xét \(I_1=\int cos2x.e^{3x}dx\) \(\Rightarrow\left\{{}\begin{matrix}u=cos2x\\dv=e^{3x}dx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=-2sin2xdx\\v=\dfrac{1}{3}e^{3x}\end{matrix}\right.\)

\(\Rightarrow I_1=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}\int sin2x.e^{3x}dx=\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\)

\(\Rightarrow I=\dfrac{1}{3}sin2x.e^{3x}-\dfrac{2}{3}\left(\dfrac{1}{3}cos2x.e^{3x}+\dfrac{2}{3}I\right)\)

\(\Rightarrow\dfrac{13}{9}I=\dfrac{1}{9}e^{3x}\left(3sin2x-2cos2x\right)\)

\(\Rightarrow I=\dfrac{1}{13}e^{3x}\left(3sin2x-2cos2x\right)+C\)

3/ \(\int e^x\left(\dfrac{1+cos2x}{2}\right)dx=\dfrac{1}{2}\int e^xdx+\dfrac{1}{2}\int cos2x.e^xdx=\dfrac{e^x}{2}+\dfrac{1}{2}I_1\)

\(I_1\) có cách tính y hệt như bài 1, bạn nguyên hàm từng phần 2 lần là xong

4/ Cũng hạ bậc tương tự câu trên và xử lý

5/ \(I=\int e^{-x}\left(\dfrac{cos3x+3cosx}{4}\right)dx=\dfrac{1}{4}\int e^{-x}\left(cos3x+3cosx\right)dx\)

\(\Rightarrow I=\dfrac{1}{4}\int e^{-x}cos3x.dx+\dfrac{3}{4}\int e^{-x}cosx.dx=I_1+I_2\)

Dùng phương pháp tương tự bài 1, lần lượt tính \(I_1\) và \(I_2\) rồi cộng vào

2/\(I=\int\dfrac{x^4}{\left(x^2-1\right)^2}dx=\int\left(1+\dfrac{2x^2-1}{\left(x^2-1\right)^2}\right)dx=\int\left(1+\dfrac{2}{x^2-1}+\dfrac{1}{\left(x^2-1\right)^2}\right)dx\)

\(=\int\left(1+\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{1}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)^2\right)dx\)

\(=\int\left(1+\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{1}{4}\left(\dfrac{1}{\left(x-1\right)^2}+\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{x+1}-\dfrac{1}{x-1}\right)\right)dx\)

\(=\int\left(1+\dfrac{3}{4}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)+\dfrac{1}{4}\dfrac{1}{\left(x+1\right)^2}+\dfrac{1}{4}\dfrac{1}{\left(x-1\right)^2}\right)dx\)

\(=x+\dfrac{3}{4}ln\left|\dfrac{x-1}{x+1}\right|-\dfrac{1}{4\left(x+1\right)}-\dfrac{1}{4\left(x-1\right)}+C\)

\(=x+\dfrac{3}{4}ln\left|\dfrac{x-1}{x+1}\right|-\dfrac{x}{2\left(x^2-1\right)}+C\)