Question

a) \(\int sin2x.cosxdx\)

b) \(\int tanxdx\)

c) \(\int\dfrac{sinx}{1+3cosx}dx\)

d) \(\int sin^3xdx\)

e) \(\int sin^2xdx\)

f) \(\int cos^23x\)

g) \(f\left(x\right)=\dfrac{1}{sin^2x.cos^2x}\)

h) \(f\left(x\right)=\dfrac{cos2x}{sin^2x.cos^2x}\)

i) \(\int2sin3x.cos2xdx\)

j) \(\int e^x\left(2+\dfrac{e^{-x}}{cos^2x}\right)dx\)

Answer 1

\(a,\int sin2x.cosxdx=\int\dfrac{1}{2}\left[sin3x+sinx\right]dx=\dfrac{1}{2}\int sin3xdx+\dfrac{1}{2}\int sinxdx=\dfrac{-1}{6}cos3x-\dfrac{1}{2}cosx\)

Answer 2

phần a bạn thêm +C vào đáp án nhé
\(i,\int2sinx3x.cos2xdx=2\int\dfrac{1}{2}\left(sin5x+sinx\right)dx=\int sin5xdx+\int sinxdx=-\dfrac{1}{5}cos5x-cosx+C\)

Answer 3

\(g,\int\dfrac{1}{sin^2x.cos^2x}=\int\dfrac{sin^2x+cos^2x}{sin^2x.cos^2x}=\int\dfrac{1}{cos^2x}dx+\int\dfrac{1}{sin^2x}dx=tanx-cotx+C\)

Answer 4

\(b,I=\int tanxdx=\int\dfrac{sinx}{cosx}dx\)
Đặt u= cosx => du= -sinxdx
\(I=\int-\dfrac{1}{u}du=-\int u^{-1}du=-ln\left|u\right|+C=-lncosx+C\)

Answer 5

\(h,f\left(x\right)=\int\dfrac{cos^2x-sin^2x}{sin^2x.cos^2x}dx=\int\dfrac{1}{sin^2x}dx-\int\dfrac{1}{cos^2x}dx=-cotx-tanx+C\)

Answer 6

\(e,\int\dfrac{1-cos2x}{2}dx=\dfrac{1}{2}\int dx-\dfrac{1}{2}\int cos2xdx=\dfrac{1}{2}x-\dfrac{1}{4}cos2x+C\)