I attempted this problem, but when I tried taking the derivative of my answer to check my work, it did not seem to match up. I cannot figure out where I went wrong. Any ideas?
These are my steps – I used half angle identities:
$$\displaystyle I=\int\cos^4x\sin^6x\ dx$$
using the identities: $\displaystyle \sin x\cos x=\frac{\sin(2x)}{2}$ and $\displaystyle \sin^2x=\frac{1-\cos(2x)}{2}$:
$$\displaystyle \int\left(\frac{\sin(2x)}{2}\right)^4\left(\frac{1-\cos(2x)}{2}\right)\ dx$$
$$\displaystyle \frac{1}{32}\int(\sin^4(2x))(1-\cos(2x))\ dx = \frac{1}{32}\int\sin^4(2x)-(\cos(2x))(\sin^4(2x))\ dx$$
We can now divide this integral into two parts: $$\displaystyle \frac{1}{32}\int\sin^4(2x)\ dx$$ and $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$
Let's integrate the first part first: $$\displaystyle \frac{1}{32}\int\sin^4(2x)\ dx$$
$$\displaystyle =\frac{1}{32}\int(1-\cos^2(2x))(\sin^2(2x)\ dx$$
$$\displaystyle =\frac{1}{32}\int(\sin^2(2x)-\sin^2(2x)(\cos^2(2x))\ dx$$
$$\displaystyle =\frac{1}{32}\int\frac{1-\cos(4x)}{2}-\frac{\sin^2(4x)}{4}\ dx$$
$$\displaystyle =\frac{1}{128}\int2-2\cos(4x)-\sin^2(4x)\ dx$$
$$\displaystyle =\frac{1}{128}\int2-2\cos(4x)-\frac{1-\cos(8x)}{2}\ dx$$
$$\displaystyle =\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{1}{256}\int1-\cos(8x)\ dx$$
$$\displaystyle =\frac{x}{64}-\frac{\sin(4x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}+C$$
Now, the second part: $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$
Let's do a u-substitution $$\displaystyle u=\sin(2x)\ \ \ \ \ \ \ \ \frac{du}{dx}=2\cos(2x)\ \ \ \ \ \ \ \ \frac12du=\cos(2x)dx$$
Now we have: $$\displaystyle \frac{-1}{64}\int u^4\ du$$
$$\displaystyle \frac{-u^5}{620} = \frac{-\sin^5(2x)}{620}+C$$
Now, putting everything together:
$$\displaystyle I=\frac{x}{64}-\frac{\sin^4(2x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{620}+C$$
$$\displaystyle =\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{620}+C$$
Correction given from answers:
Now, the second part: $$\displaystyle \frac{-1}{32}\int\cos(2x)\sin^4(2x)\ dx$$
Let's do a u-substitution $$\displaystyle u=\sin(2x)\ \ \ \ \ \ \ \ \frac{du}{dx}=2\cos(2x)\ \ \ \ \ \ \ \ \frac12du=\cos(2x)dx$$
Now we have: $$\displaystyle \frac{-1}{64}\int u^4\ du$$
$$\displaystyle \frac{-u^5}{320} = \frac{-\sin^5(2x)}{320}+C$$
Now, putting everything together:
$$\displaystyle I=\frac{x}{64}-\frac{\sin^4(2x)}{256}-\frac{x}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C$$
$$\displaystyle =\frac{3x-\sin(4x)}{256}+\frac{\sin(8x)}{2048}-\frac{\sin^5(2x)}{320}+C$$
Best Answer
Everything is correct except for some simplification right there at the bottom when you do $u^4$. Anti-derivative of $u^4$ is $u^5/5$. So,the denominator is 320 not 620.