nguyên hàm sin

\begin{aligned} &\small\text{1. Hằng đẳng thức lượng giác:}\\ & \ \ \ \ \bull sin^2x+cos^2x=1\\ & \ \ \ \ \bull \frac{1}{sin^2x}=1+cot^2x\\ & \ \ \ \ \bull \frac{1}{cos^2x}=1+tan^2x\\ &\small\text{2. Công thức cộng:}\\ & \ \ \ \ \ \bull sin(a\pm b)=sina.cosb\pm sinb.cosa\\ & \ \ \ \ \ \bull cos(a\pm b)=cosa.cosb\mp sina.cosb\\ & \ \ \ \ \ \bull tan(a\pm b)=\frac{tana \pm tanb}{1\mp tana.tanb}\\ &\small\text{3. Công thức nhân đôi:}\\ & \ \ \ \ \ \bull sin2a=2sina.cosa\\ & \ \ \ \ \ \bull cos2a=cos^2a-sin^2a=2cos^2a-1=1-2sin^2a\\ &\small\text{4. Công thức nhân ba:}\\ & \ \ \ \ \ \bull sin3a=3sina-4sin^3a\\ & \ \ \ \ \ \bull cos3a=4cos^3a-3cosa\\ &\small\text{5. Công thức hạ bậc:}\\ & \ \ \ \ \ \bull sin^2a=\frac{1-cos2a}{2}\\ & \ \ \ \ \ \bull cos^2a=\frac{1+cos2a}{2}\\ &\small\text{6.Công thức thay đổi tích trở thành tổng:}\\ & \ \ \ \ \ \bull cosa.cosb=\frac{1}{2}[cos(a-b)+cos(a+b)]\\ & \ \ \ \ \ \bull sina.sinb=\frac{1}{2}[cos(a-b)-cos(a+b)]\\ & \ \ \ \ \ \bull sina.cosb=\frac{1}{2}[sin(a-b)+sin(a+b)]\\ \end{aligned}

I=\int\frac{dx}{sin(x+a)(sin(x+b)}

\begin{aligned} &\text{Dùng tương đồng thức:}\\ &1=\frac{sin(a-b)}{sin(a-b)}=\frac{sin[(x+a)-(x+b)}{sin(a-b)}=\frac{sin(x+a)cos(x+b)-cos(x+a)sin(x+b)}{sin(a-b)}\\ &\text{Từ cơ suy ra:}\\ &I=\frac{1}{sin(a-b)}\int\frac{sin(x+a)cos(x+b)-cos(x+a)sin(x+b)}{sin(x+a)sin(x+b)}dx\\ &\ \ =\frac{1}{sin(a-b)}\int \left[ \frac{cos(x+b)}{sin(x+b)}-\frac{cos(x+a)}{sin(x+a)} \right]dx\\ &\ \ =\frac{1}{sin(a-b)}[ln|sin(x+b)|-ln|sin(x+a)|]+C \end{aligned}

\begin{aligned} &\bull J=\int\frac{dx}{cos(x+a)cos(x+b)} \text{ bởi những sử dụng tương đồng thức }1=\frac{sin(a-b)}{sin(a-b)}.\\ &\bull K=\int\frac{dx}{sin(x+a)cos(x+b)} \text{ bởi những sử dụng tương đồng thức }1=\frac{cos(a-b)}{cos(a-b)}.\\ \end{aligned}

I=\int \frac{dx}{sinx.sin\left(x+\frac{\pi}{6}\right)}

\begin{aligned} &\text{Ta có:}\\ &1=\frac{sin\frac{\pi}{6}}{sin\frac{\pi}{6}}=\frac{sin\left[\left(x+\frac{\pi}{6}\right)-x\right]}{\frac{1}{2}}=2\left[sin\left(x+\frac{\pi}{6}\right)cosx-cos\left(x+\frac{\pi}{6}\right)sinx \right]\\ &\text{Từ đó:}\\ &I=2\int\frac{\left[sin\left(x+\frac{\pi}{6}\right)cosx-cos\left(x+\frac{\pi}{6}\right)sinx \right]}{sinx.sin\left(x+\frac{\pi}{6}\right)}dx\\ &\ \ =2\int \left[\frac{cosx}{sinx}-\frac{\cos \left(x+\frac{\pi}{6}\right)}{\sin \left(x+\frac{\pi}{6}\right)} \right]dx\\ &\ \ =2\int\frac{d(sinx)}{sinx}-2\int\frac{d\left[sin\left(x+\frac{\pi}{6}\right)\right]}{sin\left(x+\frac{\pi}{6}\right)}\\ &\ \ =2ln\left|\frac{sinx}{sin\left(x+\frac{\pi}{6}\right)} \right|+C \end{aligned}

I=\int tan(x+a)tan(x+b)dx

\begin{aligned} &\text{Ta có:}\\ & tan(x+a)tan(x+b)\\ &=\frac{sin(x+a)sin(x+b)}{cos(x+a)cos(x+b)}\\ &=\frac{sin(x+a)sin(x+b)+cos(x+a)cos(x+b)}{cos(x+a)cos(x+b)}-1\\ &=\frac{cos(a-b)}{ cos(x+a)cos(x+b)}-1\\ &\text{Từ đó:}\\ &I=cos(a-b)\int\frac{dx}{cos(x+a)cos(x+b)}-1\\ &\text{Đến phía trên, tớ gặp gỡ Việc thăm dò vẹn toàn nồng độ giác ở \textbf{Dạng 1}.} \end{aligned}

\begin{aligned} &\bull J=\int cot(x+a)cot(x+b)dx\\ &\bull K=\int tan(x+a)tan(x+b)dx \end{aligned}

K=\int tan\left(x+\frac{\pi}{3}\right)cot\left(x+\frac{\pi}{6}\right)dx

\begin{aligned} &\text{Ta có:}\\ &tan\left(x+\frac{\pi}{3}\right)cot\left(x+\frac{\pi}{6}\right)\\ &=\frac{sin\left(x+\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{6}\right)}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}\\ &=\frac{sin\left(x+\frac{\pi}{3}\right)cos\left(x+\frac{\pi}{6}\right)- cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}+1\\ &=\frac{sin\left[ \left(x+\frac{\pi}{3}\right)-\left(x+\frac{\pi}{6}\right) \right]}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}+1\\ &=\frac{1}{2}.\frac{1}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}+1\\ &\text{Từ đó:}\\ &K=\frac{1}{2}\int \frac{1}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}dx+\int dx\\ &\ \ \ \ =\frac{1}{2}K_1+x+C\\ &\text{Đến phía trên, bằng phương pháp tính ở dạng 1, tớ tính được:}\\ &K_1=\int \frac{1}{cos\left(x+\frac{\pi}{3}\right)sin\left(x+\frac{\pi}{6}\right)}dx=\frac{2}{\sqrt3}ln\left| \frac{sin\left(x+\frac{\pi}{6}\right)}{cos\left(x+\frac{\pi}{3}\right)}\right|+C\\ &\text{Suy ra:}\\ &K=\frac{\sqrt3}{3}ln\left| \frac{sin\left(x+\frac{\pi}{6}\right)}{cos\left(x+\frac{\pi}{3}\right)}\right|+x+C \end{aligned}

I=\int\frac{dx}{asinx+bcosx}

\begin{aligned} &\text{Ta có:}\\ &asinx+bcosx=\sqrt{a^2+b^2} \left( \frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx\right)\\ &\Rightarrow asinx+bcosx=\sqrt{a^2+b^2}sin(x+\alpha)\\ &\Rightarrow I=\frac{1}{\sqrt{a^2+b^2}}\int \frac{dx}{sin(x+\alpha)}=\frac{1}{\sqrt{a^2+b^2}} ln \left|tan\frac{x+\alpha}{2} \right|+C \end{aligned}

I=\int\frac{2dx}{\sqrt3 sinx+cosx}

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\begin{aligned} &I=\int\frac{2dx}{\sqrt3 sinx+cosx}=\int\frac{dx}{\frac{\sqrt3}{2} sinx+\frac{1}{2}cosx}=\int \frac{dx}{sinxcos\frac{\pi}{6}+cosxsin\frac{\pi}{6}}\\ & \ \ =\int \frac{dx}{sin\left(x+\frac{\pi}{6} \right)}=\int \frac{d\left(x+\frac{\pi}{6} \right)}{sin\left(x+\frac{\pi}{6} \right)}=ln\left| tan\frac{x+\frac{\pi}{6}}{2} \right|+C=ln\left| tan\left(\frac{x}{2}+\frac{\pi}{12} \right) \right|+C \end{aligned}

I=\int\frac{dx}{asinx+bcosx}

\text{Đặt }tan\frac{x}{2}=t \Rightarrow \begin{cases}dx=\frac{2dt}{1+t^2}\\ sinx=\frac{2t}{1+t^2}\\ cosx=\frac{1-t^2}{1+t^2}\\ tanx=\frac{2t}{1-t^2} \end{cases}

K=\int\frac{dx}{sinx+tanx}

\begin{aligned} &\text{Đặt }tan\frac{x}{2}=t \Rightarrow \begin{cases}dx=\frac{2dt}{1+t^2}\\ sinx=\frac{2t}{1+t^2}\\ tanx=\frac{2t}{1-t^2} \end{cases}\\ &\text{Từ đó:}\\ &K=\int \frac{\frac{2t}{1+t^2}}{\frac{2t}{1+t^2}+\frac{2t}{1-t^2}}=\frac{1}{2}\int \frac{1-t^2}{t}dt=\frac{1}{2}\int\frac{dt}{t}-\frac{1}{2}\int tdt\\ &\ \ \ = \frac{1}{2}ln|t|-\frac{1}{4}t^2+C= \frac{1}{2}ln\left|tan\frac{x}{2}\right|-\frac{1}{4}tan^2\frac{x}{2}+C \end{aligned}

I=\int\frac{dx}{asin^2x+bsinxcosx+ccos^2x}

\begin{aligned} &I=\int\frac{dx}{(atan^2x+btanx+c)cos^2x}\\ &\text{Đặt }tanx=t\Rightarrow \frac{dx}{cos^2x}=dt\\ &\text{Suy ra: }I=\int \frac{dt}{at^2+bt+c} \end{aligned}

J=\int \frac{dx}{sin^2x-2sinxcosx-2cos^2x}

\begin{aligned} &\text{Đặt }tanx=t \Rightarrow\frac{dx}{cos^2x}=dt\\ &\Rightarrow J=\int\frac{dt}{t^2-2t-2}=\int \frac{d(t-1)}{(t-1)^2-(\sqrt3)^2}=\frac{1}{2\sqrt3}ln\left|\frac{t-1-\sqrt3}{t-1+\sqrt3} \right|+C\\ & \ \ \ \ \ \ \ \ \ =\frac{1}{2\sqrt3}ln\left|\frac{tanx-1-\sqrt3}{tanx-1+\sqrt3} \right|+C \end{aligned}

I=\int\frac{a_1sinx+b_1cosx}{a_2sinx+b_2cosx}dx

\begin{aligned} &\text{Ta thăm dò A, B sao cho:}\\ &a_1sinx+b_1cosx=A(a_2sinx+b_2cosx)+B(a_2cosx-b_2sinx) \end{aligned}

I=\int\frac{4sinx+3cosx}{sinx+2cosx}dx

\begin{aligned} &\text{Ta thăm dò A, B sao cho:}\\ &4sinx +3cosx=A(sinx+2cosx)+B(cosx-2sinx)\\ &\Rightarrow 4sinx+3cosx=(A-2B)sinx+(2A+B)cosx \Rightarrow\begin{cases} A-2B=4\\ 2A+B=3\end{cases} \Leftrightarrow\begin{cases} A=2\\B=-1\end{cases} \\ &\text{Từ đó:}\\ &I=\int\frac{2(sinx+2cosx)-(cosx-2sinx)}{sinx+2cosx}dx\\ & \ \ =2\int dx-\int \frac{d(sinx+2cosx)}{sinx+2cosx}\\ & \ \ =2x-ln|sinx+cos2x|+C \end{aligned}

I=\lmoustache sin^3x.cosx\space dx

\begin{aligned} & Ta\space có:\space sin^3x.cosxdx=\lmoustache sin^3x.d(sinx)\\ & Đặt\space u=sinx\space ta\space được:\\ & I=\lmoustache sin^3x.cosxdx=\lmoustache sin^3d(sinx)\\ & u^3du=\frac{u^4}{4}+c=\frac{sin^4x}{4}+C \end{aligned}

\intop \frac{cos^5x}{sinx}dx

\begin{aligned} & \intop \frac{cos^5x}{sinx}dx=\intop \frac{(1-sin^2x)^2dsinx}{sinx}=\intop \bigg( \frac{1}{sinx}-2sinx+sin^3x \bigg)dsinx\\ &ln|sinx|-sin^2x+\frac{sin^4x}{4}+C \end{aligned}

D=\intop \frac{dx}{3cosx+5sinx+3}

\begin{aligned} &Đặt\space tan\frac{x}{2}=t\\ &\rArr \begin{cases}dx=\frac{2dt}{1+t^2}\\sinx=\frac{2t}{1+t^2}\\ cosx=\frac{1-t^2}{1+t^2} \end{cases}\\ & Từ\space đó\space, D=\intop \frac{\frac{2dt}{1+t^2}}{3.\frac{1-t^2}{1+t^2}+5\frac{2t}{1+t^2}+3} =\frac{2dt}{3-3t^2+10+3t+2t^2}=\intop\frac{2dt}{10t+6}\\ &=\frac{1}{5}\intop \frac{d(5t+3)}{5t+3}=\frac{1}{5}ln|5t+3|+C=\frac{1}{5}ln|5tan\frac{x}{2}=3|+C\\ \end{aligned}