Question

what's the integral of cosx^3 sinx^4 from 0 to pi/4

Answer 1

\[\int\limits _0^{\pi /4}\text{Cos}[x]^3 \text{Sin}[x]^4dx=\frac{9}{280 \sqrt{2}}=0.0227284 \]

Answer 2

but what are the steps?

Answer 3

i think you would have to do repeated substituttion with u = sin[x]

Answer 4

The steps for the basic integral is revealed at the site below.

http://www.wolframalpha.com/input/?i=integrate+Cos%5Bx%5D%5E3+Sin%5Bx%5D%5E4++

Answer 5

This is how to eliminate repeated powers of sines and cosines in an integral:
http://tutorial.math.lamar.edu/Classes/CalcII/IntegralsWithTrig.aspx

You can also go to http://www.wolframalpha.com/ and type

integral of cosx^3 sinx^4

then click "show steps" to see the nitty-gritty details. WARNING -- wolframalpha does not always go the most ... efficient route. :)

Answer 6

Mathematica returns the following for the integral:\[\frac{3 \sin (x)}{64}-\frac{1}{64} \sin (3 x)-\frac{1}{320} \sin (5 x)+\frac{1}{448} \sin (7 x) \]If one applies the Mathematica "FullSimplify" function to the result above, the following is returned:\[\frac{1}{70} \sin ^5(x) (5 \cos (2 x)+9) \]\[\left\{\frac{3 \text{Sin}[x]}{64}-\frac{1}{64} \text{Sin}[3 x]-\frac{1}{320} \text{Sin}[5 x]+\frac{1}{448} \text{Sin}[7 x],\frac{1}{70} (9+5 \text{Cos}[2 x]) \text{Sin}[x]^5\right\} \]The value of the list elements above evalueated at x=0 is {0,0}. Evaluated at x=Pi/4 the following is the result:\[\left\{\frac{9}{280 \sqrt{2}},\frac{9}{280 \sqrt{2}}\right\} \]