\(=\int\left(6x^2-\dfrac{4}{x}+sin3x-cos4x+e^{2x+1}+9^{x-1}+\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\right)dx\)

\(=2x^3-4ln\left|x\right|-\dfrac{1}{3}cos3x-\dfrac{1}{4}sin4x+\dfrac{1}{2}e^{2x+1}+\dfrac{9^{x-1}}{ln9}+tanx+cotx+C\)

Biến đổi :

\(4\sin^2x+1=5\sin^2x+\cos^2x=\left(a\sin x+b\cos x\right)\left(\sqrt{3}\sin x+\cos x\right)+c\left(\sin^2x+\cos^2x\right)\)

\(=\left(a\sqrt{3}+c\right)\sin^2x+\left(a+b\sqrt{3}\right)\sin x.\cos x+\left(b+c\right)\cos^2x\)

Đồng nhấtheej số hai tử số 

\(\begin{cases}a\sqrt{3}+c=5\\a+b\sqrt{3}=0\\b+c=1\end{cases}\)

\(\Leftrightarrow\) \(\begin{cases}a=\sqrt{3}\\b=-1\\c=2\end{cases}\)

Biến đổi : 

\(5\sin x=a\left(2\sin x-\cos x+1\right)+b\left(2\cos x+\sin x\right)+c\)

         = \(\left(2a+b\right)\sin x+\left(2b-a\right)\cos x+a+c\)

Đồng nhất hệ số hai tử số : 

\(\begin{cases}2a+b=5\\2b-a=0\\a+c=0\end{cases}\)

\(\Rightarrow\) \(\begin{cases}a=2\\b=1\\c=-2\end{cases}\)

Khi đó :

\(f\left(x\right)=\frac{2\left(2\sin x-\cos x+1\right)+\left(2\cos x+\sin x\right)-2}{2\sin x-\cos x+1}\)

= \(2+\frac{2\cos x+\sin x}{2\sin x-\cos x+1}-\frac{2}{2\sin x-\cos x+1}\)

Do vậy : 

\(I=2\int dx+\int\frac{\left(2\cos x+\sin x\right)dx}{2\sin x-\cos x+1}-2\int\frac{dx}{2\sin x-\cos x+1}\)

=\(2x+\ln\left|2\sin x-\cos x+1\right|-2J+C\)

Với 

\(J=\int\frac{dx}{2\sin x-\cos x+1}\)

Ta có :

\(f\left(x\right)=\int\frac{dx}{\sqrt{3}\sin x+\cos x}=\frac{1}{2}\int\frac{dx}{\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x}=\frac{1}{2}\int\frac{dx}{\sin\left(x+\frac{\pi}{6}\right)}\)

\(=\int\frac{dx}{2\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)\cos^2\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\int\frac{dx}{\sin\left(\frac{x}{2}+\frac{\pi}{12}\right)\cos\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\int\frac{d\left(\tan\frac{x}{2}+\frac{\pi}{12}\right)}{\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)}=\ln\left|\tan\left(\frac{x}{2}+\frac{\pi}{12}\right)\right|+C\)

ko biết

a₁sinx+b₁cosx=A(a₂sinx+b₂cosx)+B(a₂cosx-b₂sinx) roi the vo ,do la dung dong nhat thuc

ma ban lam cai nay lam chi ,dai hoc dau co ma

a) \(f\left(x\right)=\sin^3x.\sin3x=\sin3x\left(\frac{3\sin x-\sin3x}{4}\right)=\frac{3}{4}\sin3x.\sin x-\frac{1}{4}\sin^23x\)

          = \(\frac{3}{8}\left(\cos2x-\cos4x\right)-\frac{1}{8}\left(1-\cos6x\right)=\frac{3}{8}\cos2x+\frac{1}{8}\cos6x-\frac{3}{8}\cos4x-\frac{1}{8}\)

Do đó : 

\(I=\int f\left(x\right)dx=\int\left(\frac{3}{8}\cos2x+\frac{1}{8}\cos6x-\frac{3}{8}\cos4x-\frac{1}{8}\right)dx=\frac{3}{16}\sin2x+\frac{1}{48}\sin6x-\frac{3}{32}\sin4x-\frac{1}{8}x+C\)

b) Ta biến đổi :

\(f\left(x\right)=\sin^3x.\cos3x+\cos^3x.\sin3x=\cos3x\left(\frac{3\sin x-\sin3x}{4}\right)+\sin3x\left(\frac{\cos3x+3\cos x}{4}\right)\)

\(=\frac{3}{4}\left(\cos3x\sin x+\sin3x\cos x\right)=\frac{3}{4}\sin4x\)

Do đó : \(I=\int f\left(x\right)dx=\frac{3}{4}\int\sin4xdx=-\frac{3}{16}\cos4x+C\)

Ta thực hiện theo các bước sau :

Bước 1 : Biến đổi 

\(a_1\sin x+b_1\cos x+c_1=A\left(a_2\sin x+b_2\cos x+c_2\right)+B\left(a_2\cos x+b_2\sin x\right)+C\)

Bước 2 : Khi đó :

\(I=\int\frac{A\left(a_2\sin x+b_2\cos x+c_2\right)+B\left(a_2\cos x+b_2\sin x\right)+C}{_2\sin x+b_2\cos x+c_2}\)

\(=A\int dx+B\int\frac{\left(a_2\cos_{ }x-b_2\sin x_{ }\right)dx}{_{ }a_2\sin x+b_2\cos x+c_2}+C\int\frac{dx}{a_2\sin x+b_2\cos x+c_2}\)

\(=Ax+B\ln\left|a_2\sin x+b_2\cos x+c_2\right|+C\int\frac{dx}{a_2\sin x+b_2\cos x+c_2}\)

Trong đó :

\(\int\frac{dx}{a_2\sin x+b_2\cos x+c_2}\)

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

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

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

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