Question

[Quyên Nguyễn]

Mọi người ơi , giúp e tính tích phân bất định với ạ ! Cảm ơn m.n ạ !

a.\(\int\frac{x+6}{\sqrt{x^2-2x+10}}dx\)

b.\(\int\frac{x}{\sqrt{3-2x-x^2}}dx\)

c.\(\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx\)

d,\(\int\frac{dx}{1+tanx}\)

e.\(\int tan^3xdx\)

f. \(\int cos^3xdx\)

g. \(\int sin^2x.cos^3xdx\)

h. \(\int sinx.cos2xdx\)

i. \(\int\frac{sin2x}{1+cos^2x}dx\)

Answer 1

[Nguyễn Việt Lâm]

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.

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

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

Answer 2

[Nguyễn Việt Lâm]

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.

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

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