First, congrats! Though you're just a 10th grade student, you know how to solve integrals.

Yes, it is more easier to solve by substituting sin x, but this method was already indicated in the first answer and I've only provided another way to solve this integral, because this is the basic ideea of providing more answers to a question.

umm... in the last answer i forgot to put + C , put the + C depends on your teacher since if the teacher is telling you to find the antiderivative, there is no need to add a C in there. Of course, if you are looking for an indefinite integral, +C is necessary. There is some confusion in the +C thing, so i didn't put it in there.

Solving this integral requires the knowledge of trigonometry and the substitution method

First, cos^5 X=cos^4 X * cos X = (1-sin^2 X)^2 * cos X

the original integral becomes

int sin^7 X * (1- sin^2 X )^2 *cos X dx

let sin X= u

it becomes

int u^7*(1-u^2)^2 du

=int u^7 (1 + u^4- 2*u^2) du

=int u^7 + u^11-2 u^9 du

by the reverse power rule

= 1/8*u^8 + 1/12*u^12- 1/5*u^10

girogiana1976's answer is also right, but since cos has a lower exponent, substituting cos with sin will be easier to do, as in the cos answer is much longer than the sin answer.

We'll solve this integral using substitution technique.

Let cos x = t => - sin x dx = dt

We'll use Pythagorean identity to write (sin x)^2, with respect to (cos x)^2:

(sin x)^2 = 1 - (cos x)^2

We'll raise to the cube both sides:

(sin x)^6 = [1 - (cos x)^2]^3

`int` (sin x)^6*(cos x)^5* sin xdx =`int` [1 - (cos x)^2]^3*(cos x)^5* sin xdx

We'll re-write the integral using the new variable t

`int` - (1-t^2)^3*t^5*dt

We'll expand the binomial:

`int` -(1 - 3t^2 + 3t^4 - t^6)*t^5*dt

`int` (-t^5 + 3t^7 - 3t^9 + t^11)dt = -t^6/6 + 3t^8/8 - 3t^10/10 + t^12/12 + C

**`int` (sin x)^7*(cos x)^5 dx = -(cos x)^6/6 + 3(cos x)^8/8 - 3(cos x)^10/10 + (cos x)^12/12 + C**