integration of trigonometric functions

• Sep 27th 2009, 10:41 PM
differentiate
integration of trigonometric functions
could someone check this for me?

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

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

THANKS
• Sep 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)

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

Originally Posted by differentiate
could someone check this for me?

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

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

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

So the integral becomes

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

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

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

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

Originally Posted by differentiate
could someone check this for me?

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

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

THANKS

Q.2 is correct.

Alternatively you can use a $\displaystyle u$ substitution.

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

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

The integral becomes

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

$\displaystyle = \int{u\,du}$

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

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

It's not difficult to show that this is equivalent to what you posted.
• Sep 27th 2009, 11:00 PM
differentiate
thank you. I get it now.

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

Originally Posted by differentiate
thank you. I get it now.

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

Correct. (Clapping)