Question

One integral question! Help asap please!
integral of cosxsin^2(2x)??

Answer 1

\[\int\limits_{}^{}cosxsin^2(2x)dx\]

Answer 2

using sin double angle identity
\[\large\int \cos x\sin ^22x=\int \cos x(2\sin x\cos x)^2=\int4\cos ^3x\sin^2x\]
it is easy to find the result of the last part

Answer 3

\[4\int\limits_{}^{}\cos^2xcosxsin^2x=4\int\limits_{}^{}\frac{ 1 }{ 2 }(1+\cos2x)\frac{ 1 }{ 2}(1-\cos2x)(\cos2x)\]

Answer 4

and then what do i do?

Answer 5

let \[t=\sin^2 x\]\[dt=\color{red}{2\cos x\sin x dx}\]

\[4\int \sin^2x\cos ^3xdx\\=2\int \sin x\cos^2 x\color{red}{2\sin x \cos xdx}\\=2\int \sin x\cos ^2x dt=2\int \sqrt{t}(1-t^2)dt\\ =2\int \sqrt{t}-t^{5/2}dt\]

Answer 6

\[\large 2(\frac{2}{3}t^\frac{3}{2}-\frac{2}{7}t^\frac{7}{2}+c)=2(\frac{2}{3}\sin^3x-\frac{2}{7}\sin^7x+c)\]

Answer 7

we suppose \[x\in(0,\pi/2)\] st\[\sin x \ge 0\]

Answer 8

line 3 of my work has a mistake ...its 1-t not 1-t^2\[\large 2\int\sin x\cos^2xdt=2\int \sqrt{t}(1-t)dt\implies 2(\frac{2}{3}t^\frac{3}{2}-\frac{2}{5}t^\frac{5}{2}+c)\\=\frac{4}{3}\sin^3x-\frac{4}{5}\sin^5x+c\]