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

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

Did you try the identity

$\displaystyle \sin^2(x) = \frac{1 - \cos(2x)}{2}$

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

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

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.

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 ?

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 $\displaystyle \frac{{\pi}}{{4}}$. 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:

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

So

$\displaystyle \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

$\displaystyle = \frac{\pi}{8} - \frac{\sin(2\pi)}{16} - (-\frac{\pi}{8}) + \frac{\sin(2\pi)}{16} = \frac{{\pi}}{{4}}$

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