Proving a trigonometric identity by integration.

Question

By considering, $$\int \sin^n (x)\cos^3(x)dx$$

Prove that: $$ \frac{ \sin^8(x)}{8} - \frac{ \sin^6(x)}{6} + \frac{1}{24} = \frac{\cos^8(x)}{8} - \frac{ \cos^6(x)}{3} + \frac{ \cos^4(x)}{4}$$

I have integrated the suggested integral successfully, that being: $$ \frac{\sin^{n+1}(x)}{n+1} - \frac { \sin^{n+3}(x)}{n+3} +c $$

I see some resemblance but I fail to connect and finish the proof. Any hints on finding the relation between the powers of sin and cos through this integral? Obviously we could expand $\sin^2(x) = (1-\cos^2(x))$ but this seems far too tedious.

Answer 1

hint...You just need to consider $$\int\sin^5x\cos^3xdx$$ either as $$\int\sin^5x(1-\sin^2x)\cos xdx$$ or as $$\int\sin x(1-\cos^2x)^2\cos^3x dx$$

These will differ by a constant which you can evaluate using eg. $x=0$

Answer 2

Use the fact that$$\sin^5(x)\cos(x)=\sin^4(x)\cos(x)\sin(x)=\bigl(1-\cos^2(x)\bigr)^2\cos(x)\sin(x)$$and do the substitution $\cos(x)=t$ and $-\sin(x)\,\mathrm dx=\mathrm dt$.

Answer 3

Hint:

$$I=\int\sin^{2m+1}x\cos^{2n+1}x\ dx=\int(1-\cos^2x)^m\cos^{2n+1}x\sin x dx$$

Set $\cos x=u$

Again,

$$I=\int(1-\sin^2x)^n\sin^{2m+1}x\cos x\ dx$$

Set $\sin x=v$

Here $2n+1=3$

Set $2m+2n+2=8,6,4$ etc.