m having hard time solving this, i keep getting the same answer over and over

• September 3rd 2012, 05:43 PM
aysar
m having hard time solving this, i keep getting the same answer over and over
the integral of sin^2 4θ dθ from -pi/4 to pi/4
• September 3rd 2012, 05:54 PM
SworD
Re: m having hard time solving this, i keep getting the same answer over and over
Did you try the identity

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$
• September 3rd 2012, 05:56 PM
aysar
Re: m having hard time solving this, i keep getting the same answer over and over
yes i did, and i got 0.785398 but my teacher says type your answer using pi
• September 3rd 2012, 05:57 PM
aysar
Re: m having hard time solving this, i keep getting the same answer over and over
wer dealing with U sub.
• September 3rd 2012, 06:01 PM
SworD
Re: m having hard time solving this, i keep getting the same answer over and over
In this case, your teacher wants an exact answer. Solve the integral without using a calculator.
• September 3rd 2012, 06:05 PM
aysar
Re: m having hard time solving this, i keep getting the same answer over and over
this is my problem , because its definite integral and how am i get an exact answer ? im just to lost. so you mean i just U sub. without calculating the integral ?
• September 3rd 2012, 06:26 PM
SworD
Re: m having hard time solving this, i keep getting the same answer over and over
There is a distinction between knowing the decimal expansion of a number, and its exact value. For example, your integral does have the value 0.785398, but that doesn't really mean anything mathematically, because its an approximation. The exact value in this case is actually $\frac{{\pi}}{{4}}$. You need to get the exact value.

One way to do that is using the identity I copied above, and then use the anti-derivative with the fundamental theorem of calculus to find the definite integral. You can use u-substitution to find the anti derivative, but it can be done otherwise:

$\sin^2(4\theta) = \frac{1 - \cos(8\theta)}{2}$

So

$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1 - \cos(8\theta)}{2} d\theta = \frac{\theta}{2} - \frac{\sin(8\theta)}{16}$ evaluated from -Pi/4 to Pi/4

$= \frac{\pi}{8} - \frac{\sin(2\pi)}{16} - (-\frac{\pi}{8}) + \frac{\sin(2\pi)}{16} = \frac{{\pi}}{{4}}$
• September 3rd 2012, 06:29 PM
aysar
Re: m having hard time solving this, i keep getting the same answer over and over
AHA !!! i got it now !! thanks my friend i really appreciate your help