\(A=\int \frac{x}{\sqrt{x+2}}dx \\ = \int \frac{x+2-2}{\sqrt{x+2}}dx \\ = \int \sqrt{x+2}-2\frac{1}{\sqrt{x+2}}dx \\ = \frac{2}{3}(x+2)^{\frac{3}{2}}-4\sqrt{x+2}+C\)

\(B=\int \frac{sinx+cosx}{\sqrt[3]{1-sin2x}}dx \\ x=\frac{\pi}{4}-u, dx=-du \\ =- \int \frac{sin(\frac{\pi}{4}-u)+cos(\frac{\pi}{4}-u)}{\sqrt[3]{1-sin(\frac{\pi}{2}-2u)}}du \\ = - \int \frac{\frac{1}{\sqrt2}cosu+\frac{1}{\sqrt2}sinu+\frac{1}{\sqrt2}cosu-\frac{1}{\sqrt2}sinu}{\sqrt[3]{1-cos2u}}du \\ = -\int \frac{\frac{2}{\sqrt2}cosu}{\sqrt[3]{1-cos2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{1-cos^2u+sin^2u}}du \\ = -\sqrt2 \int \frac{cosu}{\sqrt[3]{2sin^2u}}du \\ v=sinu, dv=cosudu \\ = -\sqrt2 \int \frac{1}{\sqrt[3]{2v^2}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} \int v^{-\frac{2}{3}}dv \\ = -\frac{\sqrt2}{\sqrt[3]2} 3v^\frac{1}{3}+C \\ = -\frac{\sqrt2}{\sqrt[3]2} 3\sqrt[3]{sin(\frac{\pi}{4}-x)}+C \)

ta có 

\(\int\frac{dx}{cosx}=\int\frac{dx}{sin\left(\frac{\Pi}{2}-x\right)}=\int\frac{dx}{2sin\left(\frac{\Pi}{4}-\frac{x}{2}\right)cos\left(\frac{\Pi}{4}-\frac{x}{2}\right)}=-\frac{1}{2}\int\frac{dsin\left(\frac{\Pi}{4}-\frac{x}{2}\right)}{2sin\left(\frac{\Pi}{4}-\frac{x}{2}\right)cos^2\left(\frac{\Pi}{4}-\frac{x}{2}\right)}=\frac{-1}{4}\int\frac{dsin\left(\frac{\Pi}{4}-\frac{x}{2}\right)}{sin\left(\frac{\Pi}{4}-\frac{x}{2}\right)\left(1-sin^2\left(\frac{\Pi}{4}-\frac{x}{2}\right)\right)}\)

đặt \(t=sin\left(\frac{\Pi}{4}-\frac{x}{2}\right)\)

ta có

\(-\frac{1}{4}\int\frac{dt}{t\left(1-t^2\right)}=-\frac{1}{4}\int\frac{dt}{t\left(1-t\right)\left(1+t\right)}=\frac{-1}{4}\int\left(\frac{1}{x}+\frac{1}{2\left(1-t\right)}-\frac{1}{2\left(1+t\right)}\right)dt=\frac{-1}{4}ln\left|x\right|+\frac{1}{8}ln\left|1+t\right|+\frac{1}{8}ln\left|1-t\right|+C\)

thay biến vào ta tìm đc tích phân cần tìm

\(\int\frac{1+sin2x+cos2x}{sinx+cosx}dx\)

\(=\int\frac{sin^2x+cos^2x+2sinxcosx+cos^2x-sin^2x}{sinx+cosx}dx\)

\(=\int\frac{\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}dx\)

\(=\int\left(sinx+cosx+cosx-sinx\right)dx=\int2cosxdx=2sinx\)

bạn tự thay cận vào nhé

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.

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

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

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

Lời giải:

\(\int ^{\frac{\pi}{2}}_{0}\frac{\sin 2x\cos x}{1+\cos x}dx=\int ^{\frac{\pi}{2}}_{0}\frac{2\sin x\cos ^2x}{\cos x+1}dx=2\int ^{\frac{\pi}{2}}_{0}\frac{\cos^2x\sin xdx}{\cos x+1}\)

\(=2\int ^{\frac{\pi}{2}}_{0}\frac{-\cos ^2xd(\cos x)}{\cos x+1}=2\int ^{0}_{1}\frac{-t^2dt}{t+1}=2\int ^{1}_{0}\frac{t^2}{t+1}dt\)

\(=2\int^1_0\frac{(t^2-1)+1}{t+1}dt=2\int ^1_0(t-1+\frac{1}{t+1})dt\)

\(=2(\frac{t^2}{2}-t+\ln|t+1|)|^{1}_0=2\ln 2-1\)

\(I=\int\frac{dx}{1+cosx}+\int\frac{sinxdx}{1+cosx}=\int\frac{d\left(\frac{x}{2}\right)}{cos^2\frac{x}{2}}-\int\frac{d\left(1+cosx\right)}{1+cosx}\)

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