Question

integrate((cosxlnsinx)/sinx)dx

Answer 1

How about a substitution? Maybe \(u = \ln(sin(x))\)?

Answer 2

could someone please help me do this question. if it was only nominator then i knew how to do it but with dinominator sinx i don't know how to solve it please help

Answer 3

u=ln(sin(x)) would work for nominator only isn't it

Answer 4

\[\int\limits \frac{ \ln \sin x }{ \sin x }\cos x~dx=\frac{ \left( \ln \sin x \right)^{2} }{ 2 }+c\]

Answer 5

Find du and your question will be answered. Please show your result.

Answer 6

Surjithayer could you please explain it to me how you got that please

Answer 7

If you find du, you can explain it to yourself.

Answer 8

\[\int\limits f \left( x \right)f \prime \left( x \right)dx=\frac{ \left( f \left( x \right) \right)^2 }{ 2 }\]

Answer 9

du=sinx/cosx
\[\int\limits_{}^{}udu=\frac{ u^2 }{ 2 }+C=\frac{ (\ln(sinx))^2 }{ 2 }+C\]

Answer 10

ok I got thank you so much for helping me with this question

Answer 11

as suggested by tkhunny
actually \[du=\frac{ \cos x }{ \sin x }dx\]

Answer 12

do you know how to solve Integrate (w^2sin(10w))dw with integration by parts

Answer 13

integrate twice.

Answer 14

\[\int\limits{w^2*\sin(10w)}dw\]

Answer 15

take w^2 as first function.
perhaps you have to integrate thhrice.

Answer 16

we have to solve it by integration by pats. what would be u and dv for this question

Answer 17

i meant integration by parts

Answer 18

\[\int\limits w^2\sin \left( 10w \right)dw=w^2\frac{ -\cos \left( 10w \right) }{10 }-\int\limits 2w \frac{ -\cos \left( 10w \right) }{ 10 }dw\]
\[=\frac{ -w^2\cos (10 w) }{ 10 }+\frac{ 1 }{ 5 }\left[ w\frac{ \sin (10w) }{ 10 }-\int\limits 1*\frac{ \sin (10 w) }{ 10 } \right]dw\]
I think now you can complete it.

Answer 19

yeah thanks

Answer 20

do yo know how to solve questions with partial pressure?

Answer 21

for example : \[\int\limits{(2x^4-x^3-9x^2+x-12)/(x^3-x^2-6x)}\]

Answer 22

i know denominator can be factored but i can't seem to factor the nominator. i don't know if i'm doing it right

Answer 23

Why do you want to factor anything?

Answer 24

\(\dfrac{2x^{4}−x^{3}−9x^{2}+x−12}{x^{3}−x^{2}−6x} = 2x + 1 + \dfrac{4x^{2} + 7x - 12}{x^{3}−x^{2}−6x}\)

At least part of that is easier.

Answer 25

\[\int\limits \frac{ 2x^4-x^3-9x^2+x-12 }{x^3-x^2-6x }dx=\int\limits \left[ 2x+1+\frac{ 4x^2+7x-12 }{ x^3-x^2-6x } \right]dx\]
\[=\frac{ 2x^2 }{ 2 }+x+\int\limits \frac{ 4x^2+7x-12 }{ x(x-3)(x+2) }dx\]
\[\frac{ x^2+7x-12 }{ x(x-3(x+2) }=\frac{? }{ x (?-3)(?+2) }+\frac{ ? }{ ?(x-3)(?+2)}+\frac{ ? }{ ?(?-3)(x+2) }\]
put x=0,in 1st ,x=3 in 2nd and x=-2 in 3rd in place of ? to make partial fractions.
\[=\frac{4* 0^2+7*0-12 }{x(0-3)(0+2) }+\frac{ 4(3)^2+7(3)-12 }{ 3(x-3)(3+2) }+\frac{ 4(-2)^2+7(-2)-12 }{ -2(-2-3)(x+2) }\]
now you can complete.