# Thread: Help me to find the solution for this math problem ?

1. ## Help me to find the solution for this math problem ?

Please help me to find the solution for this math problem ? $\int cos^3 x + sin^5 x =$

2. Originally Posted by wizard654zzz
Please help me to find the solution for this math problem ? $\int cos^3 x + sin^5 x =$
Make use of the following identities:

$\cos^3{x} = \frac{3}{4}\cos{x} + \frac{1}{4}\cos{(3x)}$

and

$\sin^5{x} = \frac{5}{8}\sin{x} - \frac{5}{16}\sin{(3x)} + \frac{1}{16}\sin{(5x)}$.

So $\int{\cos^3{x} + \sin^5{x}\,dx} = \int{\frac{3}{4}\cos{x} + \frac{1}{4}\cos{(3x)} + \frac{5}{8}\sin{x} - \frac{5}{16}\sin{(3x)} + \frac{1}{16}\sin{(5x)}\,dx}$

3. Originally Posted by wizard654zzz
Please help me to find the solution for this math problem ? $\int cos^3 x + sin^5 x =$

Put $\int \cos x(1-\sin^2\!\!x)\,dx+\int \sin^3\!\!x(1-\cos^2\!\!x)\,dx=$ $\int \cos x\,dx\,-\int \cos x\sin^2\!\!x\,dx\,+\int\sin x(1-\cos^2\!\!x)\,dx\,-\int\sin x(1-\cos^2\!\!x)\cos^2\!\!x\,dx=$

$=\int \cos x\,dx\,-\int \cos x\sin^2\!\!x\,dx\,+\int\sin x\,dx\,-\int\sin x\cos^2\!\!x\,dx\,-\int\sin x\cos^2\!\!x\,dx\,+$ $\int\sin x\cos^4\!\!x\,dx$

Now, you have integrals of the form $\int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1}$ and also immediate integrals, so you're done.

Tonio

4. Originally Posted by tonio
Put $\int \cos x(1-\sin^2\!\!x)\,dx+\int \sin^3\!\!x(1-\cos^2\!\!x)\,dx=$ $\int \cos x\,dx\,-\int \cos x\sin^2\!\!x\,dx\,+\int\sin x(1-\cos^2\!\!x)\,dx\,-\int\sin x(1-\cos^2\!\!x)\cos^2\!\!x\,dx=$

$=\int \cos x\,dx\,-\int \cos x\sin^2\!\!x\,dx\,+\int\sin x\,dx\,-\int\sin x\cos^2\!\!x\,dx\,-\int\sin x\cos^2\!\!x\,dx\,+$ $\int\sin x\cos^4\!\!x\,dx$

Now, you have integrals of the form $\int f'(x)f(x)^n\,dx=\frac{f(x)^{n+1}}{n+1}$ and also immediate integrals, so you're done.

Tonio
Mine is easier, as it doesn't involve u-substitutions.