Question

Reduction formula..
∫sin^5x cos^2x dx

Answer 1

@ganeshie8

Answer 2

@phi

Answer 3

@terenzreignz

Answer 4

sin^5x = sinx * sin^4x = sinx*(1-cos^2x)^2
helps ?

Answer 5

then what do i have to use substitution formula?

Answer 6

first expand
you will have 3 terms (3 integrals)
in each of them you will use the substitution (u=cosx)

Answer 7

note that in each of them you will have cosx to some power multiplied by sinx
since -sin is the derivative of cosx
the integrals are easy

Answer 8

okay i am gonna do it and then will write the solution to u??u are here for how many hours?

Answer 9

go for it.. even if i wont be here im sure someone else will be able to check your work

Answer 10

okay thank u.. :)

Answer 11

yw

Answer 12

https://www.wolframalpha.com/input/?i=integral+sin%28x%29^5+cos%28x%29^2

Answer 13

∫sin^5x cos^2x dx
let cosx=z
=>d/dx cosx=dz/dx
=>-sinx dx=dz
=>sinx dx=-dz
Now,
∫sin^5x cos^2x dx=∫sinx.sin^4x cos^2x dx
=∫sinx(1-cos^2x)^2. cos^2x dx
=∫sinx(1-2cos^2x+cos^4x) cos^2x dx
=∫sinx.cos^2x dx-∫sinx 2cos^2x cos^2x dx+∫sinx cos^4x cos^2x dx
=∫-z^2 dz-∫-2z^4 dz+∫-z^8 dz
=-z^2+2/2+2+2 z^4+4/4+4-z^8+8/8+8+c
=-z^4/4+2z^8/8-z^16/16+c
=-cos^4x/4+2cos^8/8-cos^16/16+c @Coolsector

Answer 14

oh god it will be cos^12/12 sorry

Answer 15

@eliassaab :( plz help

Answer 16

@UnkleRhaukus

Answer 17

@Nurali

Answer 18

\[I=\int\limits \sin ^{5}x \cos ^{2}x~dx=\int\limits \sin ^{4}x \sin x \cos ^{2}x~dx\]
\[=\int\limits \left( 1-\cos ^{2}x \right)^{2}\cos ^{2}x \sin x~dx=\int\limits \left( 1-2\cos ^{2} x+\cos ^{4}x \right)\cos ^{2}x \sin x~dx\]
\[=\int\limits \left( \cos ^{2}x-2\cos ^{4}x+\cos ^{6}x \right)dx\]
\[=\frac{ -\cos ^{3}x }{ 3 }+\frac{2 \cos ^{5}x }{ 5 }-\frac{ \cos ^{7}x }{ 7 }\]

Answer 19

correction write sin x near dx outside the bracket in last integral line.

Answer 20

oh sorry i forgot to use the formula its x^n+1/n+1

Answer 21

My head is quite hot..it's tempting !!