integration of trigonometric functions

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September 27th 2009, 10:41 PM
differentiate

integration of trigonometric functions

could someone check this for me?

1. $\int (1+cos2x)sinx\ dx$
$\int (1+cos^2x -sin^2x)sinx\ dx$
$\int (2cos^2x)sinx\ dx$
$\int 2cosx.sinx.cosx\ dx$
$\int \frac{1}{2}sin2x.cosx\ dx$
$u= \frac{1}{2}sin2x$
$\frac{du}{dx} = cosx$
$\int u\ dx$
$=\frac{ u^2}{2} = \frac{(\frac{1}{2}sin2x)^2}{2} = \frac{\frac{1}{4}sin^22x}{2} = \frac{1}{8}sin^22x + c$

2. $\int sinx.cosx\ dx$
$\int \frac{1}{2}sin2x\ dx$
$= \frac{-1}{4}cos2x + c$

THANKS
September 27th 2009, 10:48 PM
Calculus26

Once you have int( 2cos^2(x)sin(x)dx)

Let u = cos(x)

then you have simply -2 int(u^2du)

Quote:

http://www.mathhelpforum.com/math-he...1d61985b-1.gif
http://www.mathhelpforum.com/math-he...87dd5332-1.gif

isn't correct since du/dx would be cos(2x)
September 27th 2009, 10:49 PM
Prove It

Quote:

Originally Posted by differentiate

could someone check this for me?

1. $\int (1+cos2x)sinx\ dx$
$\int (1+cos^2x -sin^2x)sinx\ dx$
$\int (2cos^2x)sinx\ dx$ Correct up to here.

$= -2\int{\cos^2{x}(-\sin{x})\,dx}$

Let $u = \cos{x}$ so that $\frac{du}{dx} = -\sin{x}$ .

So the integral becomes

$-2\int{u^2\,\frac{du}{dx}\,dx}$

$= -2\int{u^2\,du}$

$= -\frac{2}{3}u^3 + C$

$= -\frac{2}{3}\cos^3{x} + C$ .
September 27th 2009, 10:52 PM
Prove It

Quote:

Originally Posted by differentiate

could someone check this for me?

1. $\int (1+cos2x)sinx\ dx$
$\int (1+cos^2x -sin^2x)sinx\ dx$
$\int (2cos^2x)sinx\ dx$
$\int 2cosx.sinx.cosx\ dx$
$\int \frac{1}{2}sin2x.cosx\ dx$
$u= \frac{1}{2}sin2x$
$\frac{du}{dx} = cosx$
$\int u\ dx$
$=\frac{ u^2}{2} = \frac{(\frac{1}{2}sin2x)^2}{2} = \frac{\frac{1}{4}sin^22x}{2} = \frac{1}{8}sin^22x + c$

2. $\int sinx.cosx\ dx$
$\int \frac{1}{2}sin2x\ dx$
$= \frac{-1}{4}cos2x + c$

THANKS

Q.2 is correct.

Alternatively you can use a $u$ substitution.

$\int{\sin{x}\cos{x}\,dx}$ .

Let $u = \sin{x}$ so that $\frac{du}{dx} = \cos{x}$ .

The integral becomes

$\int{u\,\frac{du}{dx}\,dx}$

$= \int{u\,du}$

$= \frac{1}{2}u^2 + C$

$= \frac{1}{2}\sin^2{x} + C$ .

It's not difficult to show that this is equivalent to what you posted.
September 27th 2009, 11:00 PM
differentiate

thank you. I get it now.

so for this one:
$\int cosx.sin^4x\ dx$
$u = sinx$
$\frac{du}{dx} = cosx$
$\int u^4\ du$
$= \frac{u^5}{5} + c$
$= \frac{sin^5x}{5} + c$
September 27th 2009, 11:02 PM
Prove It

Quote:

Originally Posted by differentiate

thank you. I get it now.

so for this one:
$\int cosx.sin^4x\ dx$
$u = sinx$
$\frac{du}{dx} = cosx$
$\int u^4\ du$
$= \frac{u^5}{5} + c$
$= \frac{sin^5x}{5} + c$

Correct. (Clapping)