\(sin^4x+cos^4x=\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x\)

\(=1-\frac{1}{2}\left(2sinx.cosx\right)^2=1-\frac{1}{2}sin^22x\)

\(=1-\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2}cos4x\right)=\frac{3}{4}+\frac{1}{4}cos4x\)

Khó dữ vậy trời

bài này khó quá chắc mình không giải được rồi

a) \(sin^6x+cos^6x+3sin^2x.cos^2x\)

\(=\left(sin^2x+cos^2x\right)\left(sin^4x-sin^2x.cox^2x+cos^4x\right)+3sin^2x.cos^2x\)

\(=sin^4x-sin^2x.cox^2x+cos^4x+3sin^2x.cos^2x\)

\(=sin^4x+2sin^2x.cox^2x+cos^4x=\left(sin^2x+cos^2x\right)^2=1\text{​​}\text{​}\)

b) \(sin^4x-cos^4x-\left(sinx+cosx\right)\left(sinx-cosx\right)\)

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

\(=1\left(sin^2x-cos^2x\right)-\left(sin^2x-cos^2x\right)=0\)

c) \(cos^2x+tan^2x.cos^2x\)

\(=cos^2x+\dfrac{sin^2x}{cos^2x}.cos^2x=sin^2x+cos^2x=1\)

1)

\(I=\int\left(cos^2x-cos^2x\cdot sin^3x\right)dx\\ =\int cos^2x\cdot dx-\int cos^2x\cdot sin^3x\cdot dx\\ =\frac{1}{2}\int\left(cos2x+1\right)dx+\int cos^2x\left(1-cos^2x\right)d\left(cosx\right)\\ =\frac{1}{4}sin2x+\frac{1}{2}+\frac{cos^3x}{3}-\frac{cos^5x}{5}+C\)

....

2) Xét riêng mẫu số:

\(sin2x+2\left(1+sinx+cosx\right)\\ =\left(sin2x+1\right)+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx\right)^2+2\left(sinx+cosx\right)+1\\ =\left(sinx+cosx+1\right)^2\\ =\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2\)

Khi đó:

\(I_2=\int\frac{sin\left(x-\frac{\pi}{4}\right)}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}dx\\ =-\frac{1}{\sqrt{2}}\int\frac{d\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]}{\left[\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1\right]^2}\\ =\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}cos\left(x-\frac{\pi}{4}\right)+1}+C=\frac{1}{2cos\left(x-\frac{\pi}{4}\right)+1}\)

...

lm jup mk di m.n

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

\(=\dfrac{1-3sin^2x.cos^2x-1}{1-2sin^2x.cos^2x-1}=\dfrac{3}{2}\) ko phụ thuộc x

Nên \(y'=0\) không phụ thuộc x