Đặt : t= tan\(\frac{x}{2}->dx=\frac{2dt}{1+t^2}\)

Khi đó \(I=\int\frac{4\frac{dt}{1+t^2}}{\frac{4t}{1+t^2}-\frac{1-t^2}{1+t^2}+1}=\int\frac{2dt}{t^2+2t}=\int\left(\frac{1}{t}-\frac{1}{t+2}\right)dt\)

\(ln\left|\frac{1}{t+2}\right|+C=ln\left|\frac{tan\frac{x}{2}}{tan\frac{x}{2}+2}\right|+C\)

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

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 :

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

Thực hiện theo các bước sau :

Bước 1 : Biến đổi :

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

Bước 2 : Khi đó :

\(I=\int\frac{A\left(a_2\sin x+b_2\cos x\right)+B\left(a_2\cos x-b_2\sin x\right)}{\left(a_2\sin x+b_2\cos x\right)^2}dx=A\int\frac{dx}{a_2\cos x+b_2\sin x}+B\int\frac{\left(a_2\cos x+b_2\sin x\right)dx}{\left(a_2\cos x+b_2\sin x\right)^2}\)

\(=\frac{A}{\sqrt{a^2_2+b^2_2}}\int\frac{dx}{\sin\left(x+\alpha\right)}-B\int\frac{1}{a_2\sin x+b_2\cos x}dx=\frac{A}{\sqrt{a^2_2+b^2_2}}\ln\left|\tan\left(\frac{x+\alpha}{2}\right)\right|-\frac{B}{a_2\cos x+b_2\sin x}+C\)

Trong đó : \(\sin\alpha=\frac{b_2}{\sqrt{a^2_2+b^2_2}_{ }};\cos\alpha=\frac{a_2}{\sqrt{a^2_2+b^2_2}}\)

Ta thực hiện theo các bước sau :

Bước 1 : Biến đổi

\(a_1\sin^2x+b_1\sin x\cos x+c_1\cos^2x=\left(A\sin x+B\cos x\right)\left(a_2\sin x+b_2\cos x\right)+C\left(\sin^2x+\cos^2x\right)\)

Bước 2 : Khi đó :

\(I=\int\frac{\left(A\sin x+B\cos x\right)\left(a_2\sin x+b_2\cos x\right)+C\left(\sin^2x+\cos^2x\right)}{a_2\sin x+b_2\cos x}\)

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

= \(-A\cos x+B\sin x+\sqrt{\frac{C}{a^2_a+b_2^2}}\int\frac{dx}{\sin\left(x+\alpha\right)}\)

=\(-A\cos x+B\sin x+\frac{C}{\sqrt{a_2^2+b^2_2}}\ln\left|\tan\frac{x+\alpha}{2}\right|+C\)

Trong đó :

\(\sin\alpha=\frac{b_2}{\sqrt{a_2^2}+b^{2_{ }}_2};\cos\alpha=\frac{a_2}{\sqrt{a_2^2}+b^{2_{ }}_2}\)

Ta có :

\(I=\int\frac{dx}{\left(3\tan^2x-2\tan x-1\right)\cos^2x}=\int\frac{d\left(\tan x\right)}{3\tan^2x-2\tan x-1}\)

Đặt \(t=\tan x\Rightarrow I=\int\frac{dt}{3t^2-2t-1}=\frac{1}{3}.\frac{1}{t+\frac{1}{3}}\int\left(\frac{1}{t-1}-\frac{1}{t+\frac{1}{3}}\right)dt\)

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

Thay trả lại :

\(t=\tan x\Rightarrow I=\frac{1}{4}\ln\left|\frac{3\tan x-3}{3\tan x+1}\right|+C\)

\(I=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{\cos2x+3\cos x+2}dx=\int\limits^{\frac{\pi}{2}}_0\frac{\sin x}{2\cos^2x+3\cos x+1}dx\)

Đặt \(\cos x=t\Rightarrow dt=-\sin dx\)

Với \(x=0\Rightarrow t=1\)

Với \(x=\frac{\pi}{2}\Rightarrow t=0\)

\(I=\int\limits^1_0\frac{dt}{2t^2+3t+1}=\int\limits^1_0\frac{dt}{\left(2t+1\right)\left(t+1\right)}=2\int\limits^1_0\left(\frac{1}{2t+1}+\frac{1}{2t+1}\right)dt\)

  \(=\left(\ln\frac{2t+1}{2t+1}\right)|^1_0=\ln\frac{3}{2}\)

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

b)\(\int \cos ^2 \left (\frac{x}{2}\right)dx=\int \frac{1+\cos x}{2}dx=\frac{x}{2}+\frac{\sin x}{2}+c\)

c) \(\int \frac{(2x+1)dx}{x^2+x+5}=\int \frac{d(x^2+x+5)}{x^2+x+5}=ln(x^2+x+5)+c\)

d)\(\int (2\tan x+ \cot x)^2dx=4\int \tan ^2 x+\int \cot^2 x+4\int dx=4\int \frac{1-\cos^2 x}{\cos^2 x}dx+\int \frac{1-\sin^2 x}{\sin^2 x}dx+4\int dx \)\( =4\int d(\tan x)-\int d(\cot x)-\int dx=4\tan x-\cot x-x+c\)

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:

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

\(=-\left(x+\sin^2x\right)\cos x+\int\cos x\text{dx}+2\int\sin x.\cos^2x\text{dx}\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\int\cos^2x.d\left(\cos x\right)\)

\(=-\left(x+\sin^2x\right)\cos x+\sin x-2\dfrac{\cos^3x}{3}+C\)

c) Đặt \(\left\{{}\begin{matrix}u=x+e^x\\v'=e^{2x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u'=1+e^x\\v=\dfrac{1}{2}e^{2x}\end{matrix}\right.\)

Ta có:

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

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

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

\(=\dfrac{1}{2}xe^{2x}-\dfrac{1}{4}e^{2x}+\dfrac{1}{3}e^{3x}\)