Giải:

$\int sin^2xcosxdx=\int sin^2xd(sinx)=\frac{sin^3x}{3}+c$

\(\int sin^2xcosxdx=\int sin^2xd\left(sinx\right)=\dfrac{sin^3x}{3}+c\)

8.

\(I=\int sinx.cos2xdx=\int\left(2cos^2x-1\right)sinxdx\)

\(=\int\left(1-2cos^2x\right)d\left(cosx\right)=cosx-\frac{2}{3}cos^3x+C\)

9.

\(I=\int\frac{sin2x}{1+cos^2x}dx=-\int\frac{2\left(-sinx\right).cosx}{1+cos^2x}dx=-\int\frac{d\left(cos^2x\right)}{1+cos^2x}\)

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

6.

\(I=\int cos^3xdx=\int\left(1-sin^2x\right)cosxdx\)

\(=\int\left(1-sin^2x\right)d\left(sinx\right)=sinx-\frac{1}{3}sin^3x+C\)

7.

\(I=\int sin^2x.cos^3xdx=\int sin^2x\left(1-sin^2x\right)cosxdx\)

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

x=4096

cảm ơn bạn

Câu này mình sót +3x ở sau cùng vế phải. Không hiểu vì sao đánh rồi mà lại bị mất

HELPPPP ME!!!

\(\int f\left(4x\right)dx=\frac{1}{4}\int f\left(4x\right)d\left(4x\right)=\frac{1}{16}\left(4x\right)^2+\frac{3}{4}\left(4x\right)+C\)

\(\Rightarrow\int f\left(4x\right)d\left(4x\right)=\frac{1}{4}\left(4x\right)^2+3.\left(4x\right)+C\)

\(\Rightarrow\int f\left(x+2\right)dx=\int f\left(x+2\right)d\left(x+2\right)=\frac{1}{4}\left(x+2\right)^2+3\left(x+2\right)+C\)

\(=\frac{1}{4}x^2+4x+C\)

mình cám ơn nhiều nha