Trigonometry and Integration (Calculus 2)?

Question

Having a little trouble with this one. Not sure I've done it right, maybe someone can help me out. I would deeply appreciate it.

Evaluate the following integral:
int=integral
int[sin^3xcos^x dx] = ?
also same as int[(sinx)^3(cosx)^4 dx] = ?

Thanks in advance!!

Answer 1

It can be helpful here to express this in terms of complex exponentiation. Note that sin x = (e^(ix) - e^(-ix))/(2i) and cos x = (e^(ix) + e^(-ix))/2. So:

(sin x)^3 * (cos x)^4
(e^(ix) - e^(-ix))^3/(2i)^3 * (e^(ix) + e^(-ix))^4/2^4
(e^(3ix) - 3e^(ix) + 3e^(-ix) - e^(-3ix))/(-8i) * (e^(4ix) + 4e^(2ix) + 6 + 4e^(-2ix) + e^(-4ix))/16
-1/(128i) * (e^(7ix) - e^(-7ix) + e^(5ix) - e^(-5ix) - 3e^(3ix) + 3e^(-3ix) - 3e^(ix) + 3e^(-ix))
-1/64 (sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x))

Thus, all that tedious mucking about in the complex numbers has allowed us to reduce a hard-to-integrate product of powers of trigonometric functions into a sum of very to simple integrate functions. Thus:

∫sin^3 x cos^4x dx
-1/64∫sin (7x) + sin (5x) - 3 sin (3x) - 3 sin (x) dx
-1/64 (- cos (7x)/7 - cos (5x)/5 + cos (3x) + 3 cos x) + C
cos (7x)/448 + cos (5x)/320 - cos (3x)/64 - 3 cos x/64 + C

Answer 2

Which do you choose differentiate or combine? You suggested combine on your placed up yet derive on your question??? i won't be able to answer in case you do no longer make up your techniques. :) basically in case you choose the derivation such as you asked. x^2*arctan(x^2) is derived utilising product and chain rule. Like this. d1*f2+d2(utilising chain rule)*f1 subsequently 2x*arctan(x^2)+(a million/(a million+x^4)*x^2) for the 1st one the rest artwork plenty the comparable way till you choose the integrals. if it incredibly is the case permit me comprehend.

Answer 3

ehhh, been to long, all i could help you with is i know the integral of sign would be -cos and cos would be sin

sorry but i dont remember all the algebra/calc that goes along to finish it.

Answer 4

.020408(sin(x)^2+.4)^2(cos(x))^10